Sternberg Linearization Theorem for Skew Products

Abstract

We present a special kind of normalization theorem: linearization theorem for skew products. The normal form is a skew product again, with the fiber maps linear. It appears that even in the smooth case, the conjugacy is only Hölder continuous with respect to the base. The normalization theorem mentioned above may be applied to perturbations of skew products and to the study of new persistent properties of attractors.

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Acknowledgments

We are very thankful to Ilya Schurov and Stas Minkov for their attentive reading of the preliminary versions of this article and their remarks on the presentation. We are also grateful to the Referee for numerous valuable comments. Olga Romaskevich is mainly supported by UMPA ENS Lyon. (UMR 5669 CNRS) and the LABEX MILYON (ANR-10-LABX-0070) of Université de Lyon, within the program “Investissements d’Avenir” (ANR-11-IDEX-0007) operated by the French National Research Agency (ANR) as well as by the French-Russian Poncelet laboratory (UMI 2615 of CNRS and Independent University of Moscow). Both authors are supported by RFBR project 16-01-00748.

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Correspondence to Yulij Ilyashenko.

Appendix: Proof of Theorem 3 and Other Calculations

Appendix: Proof of Theorem 3 and Other Calculations

In the appendix, we will prove the technical propositions stated above.

Hölder Properties of Some Auxiliary Functions

First let us prove

Proposition 2

Function π n (b) defined as a product of λ b in the first n points of the orbit of a linear diffeomorphism A, see Eq. 22 , is Hölder with the exponent α and

$$\|{\Pi}_{n}\|_{[\alpha]} \leq \frac{C_{\lambda} \theta^{n}}{(\mu^{\alpha}-1)q} $$

where C λ is Hölder constant for λ, 𝜃 is defined in Eq. 23 . Here and below α is given by Eq. 9 and μ is the largest magnitude of eigenvalues of A.

Proof of Proposition 2

$$\begin{array}{@{}rcl@{}} \left| {\Pi}_{n}(b_{1})-{\Pi}_{n}(b_{2}) \right| &=& \left| \prod\limits_{k=0}^{n-1} \lambda_{A^{k} b_{1}}-\prod\limits_{k=0}^{n-1} \lambda_{A^{k} b_{2}} \right|= \left| \lambda_{b_{1}}-\lambda_{b_{2}} \right| \times \left| \prod\limits_{k=1}^{n-1} \lambda_{A^{k} b_{1}} \right|\\&&+ \left|\lambda_{b_{2}}\right| \left| {\Pi}_{n-1}(A b_{1}) - {\Pi}_{n-1} (A b_{2}) \right| \leq \\ &\leq& q^{n-1} C_{\lambda} \sum\limits_{k=0}^{n-1} \left|\left|A^{k} b_{1}-A^{k} b_{2} \right|\right|^{\alpha} \leq q^{n-1} C_{\lambda} \frac{\mu^{n \alpha}-1}{\mu^{\alpha}-1} \|b_{1}-b_{2}\|^{\alpha} \\&\leq& \frac{C_{\lambda} \theta^{n}}{(\mu^{\alpha}-1)q} \|b_{1}-b_{2}\|^{\alpha} \end{array} $$

Proposition 3

Function P n (b) defined by \(P_{n}(b):=\frac {{\Pi }_{n}(b)}{\lambda _{A^{n} b}}\) is Hölder with exponent α and

$$\|P_{n}\|_{[\alpha]} \leq D^{2} C_{\lambda} B \theta^{n} $$

where B depends only on the initial map, the precise formula for B is given below, see Eq. 51.

Proof

$$\begin{array}{@{}rcl@{}} \left| P_{n}(b_{1})-P_{n}(b_{2}) \right| &=& \left| \frac{{\Pi}_{n}(b_{1}) \lambda_{A^{n} b_{2}}-{\Pi}_{n}(b_{2}) \lambda_{A^{n} b_{1}}}{\lambda_{A^{n} b_{1}} \lambda_{A^{n} b_{2}}} \right| \\ &\leq& D^{2} \left| \lambda_{A^{n} b_{2}}\prod\limits_{k=0}^{n-1} \lambda_{A^{k} b_{1}} \right.- \left. \lambda_{A^{n} b_{1}}\prod\limits_{k=0}^{n-1} \lambda_{A^{k} b_{2}} \right|\\ \end{array} $$
$$\begin{array}{@{}rcl@{}} &=& D^{2} \left| \left( \lambda_{A^{n} b_{2}}-\lambda_{A^{n} b_{1}}\right) {\Pi}_{n}(b_{1})+{\Pi}_{n+1}(b_{1}) \right. \\&&-\left.\left( \lambda_{A^{n} b_{1}} -\lambda_{A^{n} b_{2}}\right) {\Pi}_{n}(b_{2})-{\Pi}_{n+1}(b_{2})) \right| \\ &\leq&|{\Pi}_{n+1} (b_{1})-{\Pi}_{n+1}(b_{2})|+\left|\lambda_{A^{n} b_{1}}-\lambda_{A^{n} b_{2}}| |{\Pi}_{n}(b_{1})-{\Pi}_{n}(b_{2}) \right| \\&\leq& D^{2} \left[ \frac{C_{\lambda} \theta^{n+1}}{(\mu^{\alpha}-1)q}+2 q^{n} C_{\lambda} \mu^{n \alpha} \right] \|b_{1}-b_{2}\|^{\alpha} \leq D^{2} C_{\lambda} B \theta^{n} \|b_{1}-b_{2}\|^{\alpha} \end{array} $$

where B does not depend on anything but initial skew product:

$$ B(\theta,\mu,\alpha,q)=\frac{\theta}{(\mu^{\alpha}-1)q}+2 $$
(51)

Proposition 4

Function 𝜃 2,k (b 1 ,b 2 ) defined as 𝜃 2,k (b 1 ,b 2 )=|P k (b 2 )||Q k, 1 −Q k,2 | is Hölder with α as exponent and

$$\|\theta_{2,k}\|_{[\alpha]} \leq \theta^{k} D \left( C_{Q} + q^{k-1} \textup{Lip}_{x} Q \frac{C_{\lambda}}{\mu^{\alpha}-1}\right) $$

Here, \(Q_{k,1}=Q \circ {F_{0}^{k}}(b_{1},x)\) and \(Q_{k,2}=Q \circ {F_{0}^{k}}(b_{2},x)\) , and the definition of P k (b) was reminded in Proposition 2 above.

Proof

We use the results of Proposition 2 in the following chain of inequalities.

$$\begin{array}{@{}rcl@{}} \theta_{2,k} &\leq& q^{k} D \left| Q_{A^{k} b_{1}}({\Pi}_{k}(b_{1})x)-Q_{A^{k} b_{2}}({\Pi}_{k}(b_{2})x)\right| \\&\leq& q^{k} D \left| Q_{A^{k} b_{1}}({\Pi}_{k}(b_{1})x)-Q_{A^{k} b_{2}}({\Pi}_{k}(b_{1})x)\right|\\&&+q^{k} D \left| Q_{A^{k} b_{2}}({\Pi}_{k}(b_{2})x)-Q_{A^{k} b_{2}}({\Pi}_{k}(b_{1})x) \right| \\&\leq& q^{k} D C_{Q} \mu^{k \alpha} \|b_{1}-b_{2}\|^{\alpha} +q^{k} D \text{Lip}_{x} Q \|{\Pi}_{k}\|_{\mathcal{H}^{\alpha}}\|b_{1}-b_{2}\|^{\alpha} \\&\leq&\theta^{k} D \left( C_{Q} + q^{k-1} \text{Lip}_{x} Q \frac{C_{\lambda}}{\mu^{\alpha}-1}\right) \|b_{1}-b_{2}\|^{\alpha} \end{array} $$
(52)

Proposition 5

Operator L is bounded in the Lipschitz norm: there exists a constant Lip x L such that for any \(h \in \mathcal {M}\) holds

$$\textup{Lip}_{x} \left( Lh \right) \leq \textup{Lip}_{x} L \times \textup{Lip}_{x} h. $$

Moreover, if h(b,⋅)∈C l for any \(b \in \mathbb {T}^{d}\) , then Lh has the same smoothness as well and

$$\left\|(Lh)^{(l)}\right\|_{C} \leq C_{k}(L) \left\|h^{(l)}\right\|_{C}. $$

Proof

Using the explicit formula for the solution (24), as well as bounds (29) and (35), we have:

$$\begin{array}{@{}rcl@{}} \sup\limits_{x,y \in [0,1]} \left| \frac{L h (b,x)-Lh(b,y)}{x-y}\right|&=&\sup\limits_{x,y \in [0,1]} \left| \sum\limits_{k=0}^{\infty} P_{k}(b) \frac{h \circ {F_{0}^{k}}(b,x)-h \circ {F_{0}^{k}}(b,y)}{x-y} \right| \\ \end{array} $$
$$\begin{array}{@{}rcl@{}} &\leq& \sup\limits_{x,y \in [0,1]} \sum\limits_{k=0}^{\infty} P_{k}(b)\frac{\text{Lip}_{x} h |{\Pi}_{k}(b) x -{\Pi}_{k}(b) y|}{|x-y|} \\&=& \text{Lip}_{x} h \frac{D}{1-q^{2}}. \end{array} $$

The bounds for the derivatives are obtained analogously by differentiating term by term the series (24):

$$(Lh)^{(l)} =-\sum\limits_{k=0}^{\infty} P_{k}(b) {{\Pi}_{k}^{l}}(b) h^{(l)} \circ {F_{0}^{k}}. $$

Therefore,

$$\left\|(Lh)^{(l)}\right\|_{C} \leq \frac{D}{1-q^{l+1}}\left\|h^{(l)}\right\|_{C} . $$

Proposition 6

There exist polynomials T 3 (ε) and \({T_{4}^{0}}(\varepsilon )\) of degrees respectively 3 and 4 such that \({T_{4}^{0}}(0)=0\) and for any h∈N holds

$$ \textup{Lip}_{x,\varepsilon} [{\Phi} h] \leq T_{3}(\varepsilon)+{T_{4}^{0}}(\varepsilon) A_{\textnormal{Lip}} $$
(53)

Proof of Proposition 6

The proof of this proposition deals with an expression for Lip x, ε Φh which is given by

$$\sup\limits_{x,y \in [0,\varepsilon] }\frac{1}{|x-y|} \left| (1+xh_{b}(x))^{2} Q(b, x+\bar{h}_{b}(x))-(1+yh_{b}(y))^{2} Q(b, y+\bar{h}_{b}(y)) \right| $$

Since in this proposition the base coordinate b is fixed and x is changing we will permit to ourselves not to write the b index and just suppose that \(Q(x)=Q(b, x+\bar {h}_{b}(x))\) as well as h(x) = h b (x). The bound is a triangle inequality formula:

$$\begin{array}{@{}rcl@{}} \text{Lip}_{x,\varepsilon} {\Phi} h &\leq& \sup_{x,y \in [0,\varepsilon]} \left| \frac{Q(x) -Q(y)}{x-y} \right|+2 \sup_{x,y \in [0,\varepsilon]} \left| \frac{x h(x) Q(x) -y h(y) Q(y)}{x-y} \right|\\&&+ \sup_{x,y \in [0,\varepsilon]} \left| \frac{x^{2} h(x) Q(x)-y^{2} h(y) Q(y)}{x-y} \right| \leq \text{Lip}_{x} Q (1+\text{Lip}_{x,\varepsilon} \bar{h})\\ &&+ 2 \sup_{x,y \in [0,\varepsilon]} \left| \frac{x h(x) \left( Q(x)-Q(y)\right)}{x-y} \right| + 2 \sup_{x,y \in [0,\varepsilon]} \left| \frac{x Q(y) (h(x)-h(y))}{x-y} \right| \\&&+2 \sup_{y \in [0,\varepsilon]}\left| Q(y) h(y) \right|+\sup_{x,y \in [0,\varepsilon]} \left| \frac{x^{2} h(x) \left( Q(x)-Q(y)\right)}{x-y} \right|\\&&+ \sup_{x,y \in [0,\varepsilon]} \left| \frac{Q(y) \left[ x^{2}\left( h(x)-h(y)\right)+(x^{2}-y^{2})h(y)\right]}{x-y} \right| \\&\leq& \text{Lip}_{x} Q (1+\text{Lip}_{x,\varepsilon} \bar{h})+2 \varepsilon A_{C} \text{Lip}_{x} Q \text{Lip}_{x,\varepsilon} \bar{h}+ 2 \varepsilon \|Q\|_{C} A_{\text{Lip}} \\&&+ 2 \|Q\|_{C} A_{C} + \varepsilon^{2} A_{C} \text{Lip}_{x} Q \text{Lip}_{x,\varepsilon} \bar{h}+\|Q\|_{C} (\varepsilon^{2} A_{\text{Lip}}+ 2 \varepsilon A_{C}) \end{array} $$
(54)

Let us note that

$$\begin{array}{@{}rcl@{}} \text{Lip}_{x,\varepsilon} \bar{h}&=&\sup_{x,y \in [0,\varepsilon]} \left| \frac{x^{2} h(x)-y^{2} h(y)}{x-y} \right| \\&\leq& \sup_{x,y \in [0,\varepsilon]} \left|\frac{x^{2} (h(x)-h(y))}{x-y}\right|+\sup_{x,y \in [0,\varepsilon]} \left| \frac{h(y) (x^{2}-y^{2})}{x-y}\right| \\&\leq& A_{\text{Lip}} \varepsilon^{2} + A_{C} 2 \varepsilon \end{array} $$
(55)

After substitution of \(\text {Lip}_{x,\varepsilon } \bar {h}\) in Eq. 54 by Eq. 55, we have the result with

$$\begin{array}{@{}rcl@{}} T_{3}(\varepsilon)&=&2{A_{C}^{2}} \text{Lip}_{x} Q \varepsilon^{3} + 4 {A_{C}^{2}} \text{Lip}_{x} Q \varepsilon^{2}+2 A_{C} (\text{Lip}_{x} Q +\|Q\|_{C})\varepsilon+ \text{Lip}_{x} Q + 2 \|Q\|_{C} A_{C} \\ {T_{4}^{0}}(\varepsilon)&=&\text{Lip}_{x} Q A_{C} \varepsilon^{4} + 2 A_{C} \text{Lip}_{x} Q \varepsilon^{3} + \text{Lip}_{x} Q \varepsilon^{2} + 2 \|Q\|_{C} \varepsilon \end{array} $$

Now let us prove the analogous proposition for the Hölder norm of the operator Φ:

Proposition 7

If \(h \in \mathcal {H}^{\alpha }_{\varepsilon }\) then \({\Phi } h \in \mathcal {H}^{\alpha }_{\varepsilon }\) as well. And, moreover, for h∈N, there exist polynomials \(\tilde {T}_{2}(\varepsilon )\) and \(\tilde {T}_{4}^{0}(\varepsilon ), \tilde {T}_{4}^{0}(\varepsilon )(0)=0\) of degrees 2 and 4 correspondingly such that

$$\left\|{\Phi} h \right\|_{[\alpha], \varepsilon} \leq \tilde{T}_{4}^{0}(\varepsilon) A_{\alpha}+ \tilde{T}_{2}(\varepsilon) $$

Proof

To estimate Hölder norm of the shift operator, we need some more triangle inequalities.

$$\begin{array}{@{}rcl@{}} |{\Phi} h(b_{1}, x)-{\Phi} h (b_{2},x)|&=&|(1+x h_{b_{1}})^{2} Q_{b_{1}} - (1+xh_{b_{2}})^{2} Q_{b_{2}}| \\&\leq& |Q_{b_{1}}-Q_{b_{2}}|+ 2 x \left|h_{b_{1}} Q_{b_{1}}-h_{b_{2}} Q_{b_{2}}\right|\\&&+ x^{2} \left| h_{b_{1}}^{2} Q_{b_{1}} - h_{b_{2}}^{2} Q_{b_{2}} \right| \\ &\leq& \left| Q(b_{1}, x+\bar{h}_{b_{1}})-Q(b_{1}, x+\bar{h}_{b_{2}}) \right|\\&&+ \left| Q(b_{1}, x+\bar{h}_{b_{2}})-Q(b_{2}, x+\bar{h}_{b_{2}}) \right| \\&&+ 2 \varepsilon \left| h_{b_{1}} \left( Q(b_{1}, x+\bar{h}_{b_{1}})-Q(b_{1}, x+\bar{h}_{b_{2}})\right) \right|\\&&+ 2 \varepsilon \left| h_{b_{1}} \left( Q(b_{1}, x+\bar{h}_{b_{2}}) -Q(b_{2}, x+\bar{h}_{b_{2}}) \right) \right| \\&&+ 2 \varepsilon \left| Q_{b_{2}} (h_{b_{1}}-h_{b_{2}}) \right|+\varepsilon^{2} h_{b_{1}}^{2} \left| Q(b_{1}, x+\bar{h}_{b_{1}}) - Q(b_{1}, x+\bar{h}_{b_{2}}) \right|\\&&+ \varepsilon^{2} h_{b_{1}}^{2} \left| Q(b_{1}, x+\bar{h}_{b_{2}})-Q(b_{2}, x+\bar{h}_{b_{2}}) \right|\\&&+\varepsilon^{2} |Q_{b_{2}}| \left| h_{b_{1}}^{2}-h_{b_{2}}^{2} \right| \\&\leq& \|b_{1}-b_{2}\|^{\alpha} \left( \tilde{T}_{4}^{0}(\varepsilon) A_{\alpha}+\tilde{T}_{2}(\varepsilon) \right) \end{array} $$

where

$$\tilde{T_{4}}^{0} (\varepsilon)=\text{Lip}_{x} Q \varepsilon^{2} + 2 \varepsilon^{3} A \text{Lip}_{x} Q + 2 \varepsilon \|Q\|_{C} + \varepsilon^{4} A^{2} \text{Lip}_{Q} + 2 \|Q\|_{C} A \varepsilon^{2} $$

and

$$\tilde{T}_{2}(\varepsilon)=C_{Q}+ 2 \varepsilon A C_{Q}+ \varepsilon^{2} A^{2} C_{Q} $$

All the propositions stated above are proven. This completes the proof of our main result – Theorem 2.

Now we are ready to prove that the conjugacy is smooth in fiber variable, and Hölder with its derivatives in base variables.

Proof of Theorem 3

The proof of a smooth version of Theorem 2 is analogous to the proof of the latter theorem. Here, we will give a sketch of the proof: we will only show that the conjugacy H is (k−2)– smooth with respect to the fiber variable. The proof of the fact that its fiber derivatives are now Hölder on b is analogous to the proof of the Hölder property fot the function H itself and we do not give it here.

The idea is to change the space \(\mathcal {N}\) in an appropriate way. For some constants A 0,…, A l > 0 and κ 0,…, κ l > 0 let us define the space

$$\mathcal{N}_{l}:=\left\{ h(\cdot, b) \in C^{l}([0,\varepsilon]) : \|h\|_{C} \leq A_{0}, {\ldots} , \left\|h^{(l)}\right\|_{C} \leq A_{l} \right\} $$

with the norm

$$ \|h\|_{*}=\kappa_{0} \|h\|_{C} + {\ldots} + \kappa_{l} \|h^{(l)}\|_{C}. $$
(56)

We have now to prove the analogues of Lemmas 1, 2, and 3 above, and then follow the argument in Theorem 2. The homological and shift operators will stay the same although the functional spaces in which they act will be smaller, and the metric will be not continuous but a smooth one. □

Lemma 1 (Smooth case)

Operator L is bounded in the norm ( 56 ).

Proof

$$\begin{array}{@{}rcl@{}} \|L h\|_{*} &=& \sum\limits_{j=0}^{l} \kappa_{j} \| (L h)^{(j)}\|_{C} \leq \sum\limits_{j=0}^{l} \frac{D \kappa_{j}}{1-q^{j+1}}\|h^{(j)}\|_{C} \\&\leq&\frac{D}{1-q} \sum\limits_{j=0}^{\infty} \kappa_{j} \|h^{(j)}\|_{C} = \frac{D}{1-q}\|h\|_{*} \end{array} $$
(57)

For the space \(\mathcal {N}_{l}\) to map to itself by LΦ, we should choose constants A 0, A 1,…, A l appropriately. For LΦ to be contracting in the space, we should appropriately choose κ 0,…, κ l . Let us show that these two choices can be made without complications and the analogues of Lemmas 2 and 3 hold.

In what concerns the operator Φ, we will use its presentation (43) and calculate the derivatives for k = 0,…, l: by the Leibnitz rule:

$$({\Phi} h)^{(k)}=\sum\limits_{j=0}^{k} {C_{k}^{j}} ((1+x h(b,x))^{2})^{(j)}Q^{(k-j)}(b, x+\bar{h}) $$

The explicit form of the right-hand side is not as important as a fact that it can be written as a sum of polynomials in derivatives of \(h, \bar {h}\) and Q. Indeed, there exist polynomials τ 0,…, τ l and σ 0,…, σ l such that

$$\begin{array}{@{}rcl@{}} ({\Phi} h)^{(k)}&=&\sum\limits_{j=0}^{k} {C_{k}^{j}} \tau_{j}(x,h, \ldots, h^{(j)}) \sigma_{j} \\&&\times\left( x, \bar{h}, \ldots, \bar{h}^{(k-j)}, Q\left( b, x+\bar{h}), \ldots, \right.\right. \left. \left. Q^{(k-j)}(b,x+\bar{h} \right)\right) \end{array} $$
(58)

We will estimate the continuous norm of the right-hand side of Eq. 58 in \(\mathbb {T}^{d} \times [0,\varepsilon ]\). So, we will have that for some polynomials T j and S j there is a bound

$$\begin{array}{@{}rcl@{}} \|L{\Phi} h^{(k)}\|_{C, \varepsilon} &\leq& \sum\limits_{j=0}^{k} {C_{k}^{j}} T_{j}(\varepsilon, A_{0}, \ldots, A_{j}) S_{j}\\&& \times\left( \varepsilon, A_{0}, \ldots, A_{k-j}, \|Q\|_{C}, \ldots, \right. \left. \|Q^{(k-j)}\|_{C}\right) \end{array} $$
(59)

Note that the coefficient in front of A k in this expression is a polynomial that has no free term. Indeed, A k comes only from T k or S 0: in both of the cases, A k is multiplied by at least one ε. Hence, we need to find A 0,…A l such that l+1 equations hold for some polynomials \(P_{k}, {P_{k}^{0}}, {P_{k}^{0}}(0)=0\):

$$ {P_{k}^{0}}(\varepsilon) A_{k}+P_{k}(\varepsilon) C(A_{0}, \ldots, A_{k-1}) \leq A_{k}, k=0, \ldots, l $$
(60)

First, we take ε such that all polynomials \({P_{k}^{0}}(\varepsilon )<1\). Then we satisfy the Eq. 60 one by one, starting from k = 0 by choosing A k one by one, starting with A 0 and by increasing the index.

Now we have to prove that operator LΦ is contracting in the space \(\mathcal {N}\) if κ 0,…κ l are properly chosen.

One can show that

$$\begin{array}{@{}rcl@{}} \|L {\Phi} h - L {\Phi} g\|_{*} &\leq& \frac{D}{1-q} \sum\limits_{j=0}^{l} \kappa_{j} \| \sum\limits_{k=0}^{j} {C_{j}^{k}} \\&&\times\left( \tau_{k}(x,h, {\ldots} h^{(k)}) \sigma_{k} \left( x, \bar{h}, \ldots, \bar{h}^{(j-k)}, Q(b,x+h), \ldots, \right. \right. \\&&\left. \left. Q^{(j-k)}(b,x+h)\right)-\right.\left. \tau_{k}\left( x,g, {\ldots} g^{(k)}) \sigma_{k}(x, \bar{g}, \ldots, \bar{g}^{(j-k)}, Q(b,x+g), \ldots,\right.\right. \\&&\left.\left. Q^{(j-k)}(b,x+g)\right) \right) \|_{C, \varepsilon} \leq \sum\limits_{j=0}^{l} \|h^{(k)}-g^{(k)}\|_{C, \varepsilon}\\&&\times \left( \kappa_{k} {U_{k}^{0}}(\varepsilon)+\sum\limits_{j={k+1}}^{l} U_{j}(\varepsilon) \kappa_{j} \right) \end{array} $$

for some polynomials \(U_{j}, {U_{j}^{0}}\). For the right-hand side to be less than ξhg for some ξ<1, the following system should be satisfied:

$$\begin{array}{@{}rcl@{}} {U_{0}^{0}} (\varepsilon)+ \frac{\kappa_{1}}{\kappa_{0}} U_{0}(\varepsilon) + {\ldots} \frac{\kappa_{l}}{\kappa_{0}} U_{0}(\varepsilon) \leq \xi\\ {U_{1}^{0}} (\varepsilon)+ \frac{\kappa_{2}}{\kappa_{1}} U_{0}(\varepsilon) + {\ldots} \frac{\kappa_{l}}{\kappa_{1}} U_{0}(\varepsilon) \leq \xi \\ {\ldots} \\ U_{l-1}^{0} (\varepsilon) + \frac{\kappa_{l}}{\kappa_{l-1}} U_{l-1}(\varepsilon) \leq \xi\\{U_{l}^{0}}(\varepsilon) \leq \xi \end{array} $$

One can choose ε in such a way that \({U^{0}_{k}}(\varepsilon )<\xi \). Then, the last inequality in the list is true, by taking any κ l and κ l−1 big enough, we satisfy the before-last inequality, and we proceed in satisfying these inequalities from the last one till the first one.

So, we obtain a contracting operator. We have proved that in the space \(\mathcal {N}_{l}\) of functions defined in a neighborhood of the base with a metric chosen appropriately, there is a contracting operator LΦ. Its fixed point is the needed conjugacy which will be sufficiently smooth on the fiber variable x.

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Ilyashenko, Y., Romaskevich, O. Sternberg Linearization Theorem for Skew Products. J Dyn Control Syst 22, 595–614 (2016). https://doi.org/10.1007/s10883-016-9319-6

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Keywords

  • Normal forms
  • Skew products
  • Sternberg theorem

Mathematics Subject Classification (2010)

  • 37G05
  • 37D30