In this chapter, we give either complete proofs or schemes of proof of the following main results:

  • If a diffeomorphism f of a smooth closed manifold has the Lipschitz shadowing property, then f is structurally stable (Theorem 2.3.1);

  • a diffeomorphism f has the Lipschitz periodic shadowing property if and only if f is Ω-stable (Theorem 2.4.1);

  • if a diffeomorphism f of class C 2 has the Hölder shadowing property on finite intervals with constants \(\mathcal{L},C,d_{0},\theta,\omega\), where θ ∈ (1∕2, 1) and θ + ω > 1, then f is structurally stable (Theorem 2.5.1);

  • there exists a homeomorphism of the interval that has the Lipschitz shadowing property and a nonisolated fixed point (Theorem 2.6.1);

  • if a vector field X has the Lipschitz shadowing property, then X is structurally stable (Theorem 2.7.1).

The structure of the chapter is as follows.

We devote Sects. 2.12.3 to the proof of Theorem 2.3.1. In Sect. 2.1, we prove theorems of Maizel’ and Pliss relating the so-called Perron property of difference equations and hyperbolicity of sequences of linear automorphisms, Sect. 2.2 is devoted to the Mañé theorem (Theorem 1.3.7), and in Sect. 2.3, we reduce the proof of Theorem 2.3.1 to results of the previous two sections.

Theorem 2.4.1 is proved in Sect. 2.4; Theorem 2.5.1 is proved in Sect. 2.5; Theorem 2.6.1 is proved in Sect. 2.6.

Finally, Sect. 2.7 is devoted to the proof of Theorem 2.7.1.

2.1 Maizel’ and Pliss Theorems

Let \(I =\{ k \in \mathbb{Z}:\; k \geq 0\}\). Let \(\mathcal{A} =\{ A_{k},\;k \in I\}\) be a sequence of linear isomorphisms

$$\displaystyle{A_{k}:\; \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}.}$$

We assume that there exists a constant N ≥ 1 such that

$$\displaystyle{ \|A_{k}\|,\|A_{k}^{-1}\| \leq N,\quad k \in I. }$$
(2.1)

We relate to this sequence two difference equations, the homogeneous one,

$$\displaystyle{ x_{k+1} = A_{k}x_{k},\quad k \in I, }$$
(2.2)

and the inhomogeneous one,

$$\displaystyle{ x_{k+1} = A_{k}x_{k} + f_{k+1},\quad k \in I. }$$
(2.3)

Definition 2.1.1

We say that the sequence \(\mathcal{A}\) has the Perron property on I if for any bounded sequence f k , Eq. (2.3) has a bounded solution.

Set

$$\displaystyle{\,F(k,l) = \left \{\begin{array}{ll} A_{k-1} \circ \ldots \circ A_{l}, &k > l; \\ \mbox{ Id}, &k = l; \\ A_{k}^{-1} \circ \ldots \circ A_{l-1}^{-1},&k < l.\\ \end{array} \right.}$$

Definition 2.1.2

We say that the sequence \(\mathcal{A}\) is hyperbolic on I if there exist constants C > 0 and λ ∈ (0, 1) and projections P k , Q k , kI, such that if \(S_{k} = P_{k}\mathbb{R}^{n}\) and \(U_{k} = Q_{k}\mathbb{R}^{n}\), then

$$\displaystyle{ S_{k} \oplus U_{k} = \mathbb{R}^{n}; }$$
(2.4)
$$\displaystyle{ A_{k}S_{k} = S_{k+1},\quad A_{k}U_{k} = U_{k+1}; }$$
(2.5)
$$\displaystyle{ \vert \,F(k,l)v\vert \leq C\lambda ^{k-l}\vert v\vert,\quad v \in S_{ l},\;k \geq l; }$$
(2.6)
$$\displaystyle{ \vert \,F(k,l)v\vert \leq C\lambda ^{l-k}\vert v\vert,\quad v \in U_{ l},\;k \leq l; }$$
(2.7)
$$\displaystyle{ \|\,P_{k}\|,\|Q_{k}\| \leq C. }$$
(2.8)

In the relations above, k, lI.

Our first main result in this section is the following statement.

Theorem 2.1.1 (Maizel’)

If the sequence \(\mathcal{A}\) has the Perron property on I, then this sequence is hyperbolic on I.

Remark 2.1.1

Of course, it is well known that a hyperbolic sequence \(\mathcal{A}\) has the Perron property on I (see Lemma 2.1.6 below), so the properties of \(\mathcal{A}\) in the above theorem are equivalent. We formulate it in the above form since this implication is what we really need (and since precisely this statement was proved by Maizel’).

Proof

Thus, we assume that the sequence \(\mathcal{A}\) has the Perron property on I.

Let us denote by \(\mathcal{B}\) the Banach space of bounded sequences x = {x k }, where \(x_{k} \in \mathbb{R}^{n}\) and kI, with the usual norm

$$\displaystyle{\|x\| =\sup _{k\in I}\vert x_{k}\vert.}$$

A sequence \(x \in \mathcal{B}\) that satisfies Eq. (2.2) (or (2.3)) will be called a \(\mathcal{B}\)-solution of the corresponding equation.

Denote

$$\displaystyle{V _{1} = \left \{x_{0}:\; x = (x_{0},x_{1},\ldots )\quad \text{is a }\mathcal{B} -\text{solution of}\quad \text{(2.2)}\right \}.}$$

Since Eq. (2.2) is linear and \(\mathcal{B}\) is a linear space, V 1 is a linear space as well. Denote by V 2 the orthogonal complement of V 1 in \(\mathbb{R}^{n}\) and by P the orthogonal projection to V 1.

The difference of any two \(\mathcal{B}\)-solutions of Eq. (2.3) with a fixed \(f \in \mathcal{B}\) is a \(\mathcal{B}\)-solution of Eq. (2.2). It is easily seen that for any \(f \in \mathcal{B}\) there exists a unique \(\mathcal{B}\)-solution of Eq. (2.3) (we denote it T( f)) such that (T( f))0V 2.

The defined operator

$$\displaystyle{T:\; \mathcal{B} \rightarrow \mathcal{B}}$$

plays an important role in the proof. Clearly, the operator T is linear.

Lemma 2.1.1

The operator T is continuous.

Proof

Since we know that the operator T is linear, it is enough to show that the graph of T is closed; then our statement follows from the closed graph theorem.

Thus, assume that

$$\displaystyle{\,f_{n} = (\,f_{0}^{n},\ldots ) \in \mathcal{B},\quad y_{ n} = (\,y_{0}^{n},\ldots ) \in \mathcal{B},}$$

y n = T( f n ), f n f, and y n y = ( y 0, ) in \(\mathcal{B}\).

Then, clearly, y 0V 2.

Fix kI and pass in the equality

$$\displaystyle{\,y_{k+1}^{n} = A_{ k}y_{k}^{n} + f_{ k+1}^{n}}$$

to the limit as n to show that

$$\displaystyle{\,y_{k+1} = A_{k}y_{k} + f_{k+1}.}$$

Hence, y = T( f), and the graph of T is closed. □

Lemma 2.1.1 implies that there exists a constant r > 0 such that

$$\displaystyle{ \|T(\,f)\| \leq r\|\,f\|,\quad f \in \mathcal{B}. }$$
(2.9)

Without loss of generality, we assume that

$$\displaystyle{ rN \geq 1, }$$
(2.10)

where N is the constant in (2.1).

Denote

$$\displaystyle{X(k) = \left \{\begin{array}{ll} F(k,0), &k > 0; \\ \mbox{ Id}, &k = 0; \\ F(0,-k),&k < 0.\\ \end{array} \right.}$$

Straightforward calculations show that the formula

$$\displaystyle{ y_{k} =\sum _{ u=0}^{k}X(k)PX(-u)f_{ u} -\sum _{u=k+1}^{\infty }X(k)(\mbox{ Id} - P)X(-u)f_{ u} }$$
(2.11)

represents a solution of Eq. (2.3) provided that the series in the second summand converges.

We can obtain a shorter variant of formula (2.11) by introducing the “Green function”

$$\displaystyle{G(k,u) = \left \{\begin{array}{ll} X(k)PX(-u), &0 \leq u \leq k;\\ - X(k)(\mbox{ Id} - P)X(-u), &0 \leq k < u. \\ \end{array} \right.}$$

Then formula (2.11) becomes

$$\displaystyle{ y_{k} =\sum _{ u=0}^{\infty }G(k,u)f_{ u}. }$$
(2.12)

Lemma 2.1.2

Let k 0, k 1, kI and let \(\xi \in \mathbb{R}^{n}\) be a nonzero vector with | ξ | ≤ 1. Then

$$\displaystyle{ \vert X(k)P\xi \vert \sum _{u=k_{0}}^{k}\vert X(u)\xi \vert ^{-1} \leq r,\quad 0 \leq k_{ 0} \leq k, }$$
(2.13)

and

$$\displaystyle{ \vert X(k)(\mathit{\mbox{ Id}} - P)\xi \vert \sum _{u=k}^{k_{1} }\vert X(u)\xi \vert ^{-1} \leq 2rN,\quad 0 \leq k \leq k_{ 1}. }$$
(2.14)

Proof

Without loss of generality, we may take f 0 = 0. Fix l 0, l 1I such that l 0l 1. Take a sequence f with f i = 0,  i > l 1. Then formula (2.12) takes the form

$$\displaystyle{\,y_{l} =\sum _{ u=0}^{l_{1} }G(l,u)f_{u}.}$$

For ll 1, all the indices u in this sum do not exceed l 1, and we apply the first line in the definition of G. Thus,

$$\displaystyle{\,y_{l} = X(l)P\sum _{u=0}^{l_{1} }X(-u)f_{u}.}$$

Hence, if ll 1, then y l is the image under X(l) of a vector from V 1 that does not depend on l. It follows that the sequence y (with the exception of a finite number of entries) is a solution of Eq. (2.2) with initial value from V 1. Hence, \(y \in \mathcal{B}\). Since f 0 = 0,

$$\displaystyle{\,y_{0} = -(\mbox{ Id} - P)\sum _{u=0}^{l_{1} }X(-u)f_{u} \in V _{2}.}$$

Thus, y = T( f), and ∥ y∥ ≤ r∥ f∥.

Now we specify the choice of f. Let x i = X(i)ξ; since ξ ≠ 0, x ≠ 0 as well. Set

$$\displaystyle{\,f_{i} = \left \{\begin{array}{ll} 0, &i < l_{0}; \\ x_{i}/\vert x_{i}\vert,&l_{0} \leq i \leq l_{1}; \\ 0, &i > l_{1}.\\ \end{array} \right.}$$

Since ∥ f∥ = 1, inequality (2.9) implies that

$$\displaystyle{ \left \vert \sum _{u=l_{0}}^{l_{1} }G(k,u)x_{i}/\vert x_{i}\vert \right \vert = \vert \,y_{l}\vert \leq r. }$$
(2.15)

We take l = l 1 = k and l 0 = k 0 in (2.15) and conclude that

$$\displaystyle{r \geq \left \vert \sum _{u=k_{0}}^{k}G(k,u)x_{ u}/\vert x_{u}\vert \right \vert = \left \vert \sum _{u=k_{0}}^{k}X(k)PX(-u)X(u)\xi /\vert X(u)\xi \vert \right \vert =}$$
$$\displaystyle{= \vert X(k)P\xi \vert \sum _{u=k_{0}}^{k}\vert X(u)\xi \vert ^{-1},}$$

which is precisely inequality (2.13).

We prove inequality (2.14) using a similar reasoning.

First we consider 0 < kk 1. We take l = k − 1, l 0 = k, and l 1 = k 1 in (2.15) and get the estimates

$$\displaystyle{r \geq \left \vert \sum _{u=k}^{k_{1} }G(k,u)x_{u}/\vert x_{u}\vert \right \vert = \left \vert \sum _{u=k}^{k_{1} }X(k - 1)(\mbox{ Id} - P)X(-u)X(u)\xi /\vert X(u)\xi \vert \right \vert =}$$
$$\displaystyle{= \vert X(k-1)(\mbox{ Id}-P)\xi \vert \sum _{u=k}^{k_{1} }\vert X(u)\xi \vert ^{-1} = \vert A_{ k-1}^{-1}X(k)(\mbox{ Id}-P)\xi \vert \sum _{ u=k}^{k_{1} }\vert X(u)\xi \vert ^{-1} \geq }$$
$$\displaystyle{\geq \| A_{k-1}\|^{-1}\vert X(k)(\mbox{ Id} - P)\xi \vert \sum _{ u=k}^{k_{1} }\vert X(u)\xi \vert ^{-1}.}$$

Applying inequality (2.1), we see that in this case,

$$\displaystyle{\vert X(k)(\mbox{ Id} - P)\xi \vert \sum _{u=k}^{k_{1} }\vert X(u)\xi \vert ^{-1} \leq rN.}$$

Now we consider 0 = k < k 1 and apply the previous estimate with k = 1:

$$\displaystyle{\vert X(0)(\mbox{ Id}-P)\xi \vert \sum _{u=0}^{k_{1} }\vert X(u)\xi \vert ^{-1} = \vert X(0)(\mbox{ Id}-P)\xi \vert \sum _{ u=1}^{k_{1} }\vert X(u)\xi \vert ^{-1}+\vert (\mbox{ Id}-P)\xi \vert \leq }$$
$$\displaystyle{\leq \| A_{0}\|^{-1}\vert X(1)(\mbox{ Id} - P)\xi \vert \sum _{ u=1}^{k_{1} }\vert X(u)\xi \vert ^{-1} + 1 \leq rN + 1 \leq 2rN}$$

(recall that | ξ | ≤ 1 and rN ≥ 1).

For k = k 1 = 0, our inequality is trivial. □

Lemma 2.1.3

Let k 0, k 1, k, sI and let \(\xi \in \mathbb{R}^{n}\) be a unit vector. Denote

$$\displaystyle{\mu = 1 - (2rN)^{-1}.}$$

Then the following inequalities are satisfied:

if Pξ ≠ 0, then

$$\displaystyle{ \sum _{u=k_{0}}^{s}\vert X(u)P\xi \vert ^{-1} \leq \mu ^{k-s}\sum _{ u=k_{0}}^{k}\vert X(u)P\xi \vert ^{-1},\quad k_{ 0} \leq s \leq k; }$$
(2.16)

if (IdP)ξ ≠ 0, then

$$\displaystyle{ \sum _{u=s}^{k_{1} }\vert X(u)(\mathit{\mbox{ Id}} - P)\xi \vert ^{-1} \leq \mu ^{s-k}\sum _{ u=k}^{k_{1} }\vert X(u)(\mathit{\mbox{ Id}} - P)\xi \vert ^{-1},\quad k \leq s \leq k_{ 1}. }$$
(2.17)

Proof

Denote

$$\displaystyle{\phi _{i} =\sum _{ u=k_{0}}^{i}\vert X(u)P\xi \vert ^{-1},\quad i \geq k_{ 0},}$$

and

$$\displaystyle{\psi _{i} =\sum _{ u=i}^{k_{1} }\vert X(u)(\mbox{ Id} - P)\xi \vert ^{-1},\quad i \leq k_{ 1}.}$$

Let us prove inequality (2.16). Since ≠ 0, ϕ i > 0. Clearly, ϕ i ϕ i−1 = | X(i) |−1. Replacing ξ by (and noting that | | ≤ 1) in (2.13), we see that

$$\displaystyle{ \frac{\phi _{i}} {\phi _{i} -\phi _{i-1}} \leq r \leq 2rN.}$$

Hence,

$$\displaystyle{(2rN)^{-1} \leq \frac{\phi _{i} -\phi _{i-1}} {\phi _{i}} = 1 -\frac{\phi _{i-1}} {\phi _{i}},}$$

and

$$\displaystyle{\phi _{i-1} \leq (1 - (2rN)^{-1})\phi _{ i}.}$$

Iterating this inequality, we conclude that

$$\displaystyle{\phi _{s} \leq (1 - (2rN)^{-1})^{k-s}\phi _{ k},\quad k \geq s.}$$

We prove inequality (2.17) similarly. We note that ψ i > 0 and that ψ i ψ i+1 = | X(i)(Id − P)ξ |−1. After that, we replace ξ by (Id − P)ξ in (2.14) and show that

$$\displaystyle{\psi _{i+1} \leq (1 - (2rN)^{-1})\psi _{ i}.}$$

Iterating this inequality, we get (2.17). □

Now we prove that the sequence \(\mathcal{A}\) is hyperbolic.

Lemma 2.1.4

The following inequalities are satisfied:

$$\displaystyle{\|X(k)PX(-s)\| \leq r^{2}\mu ^{k-s},\quad 0 \leq s \leq k,}$$

and

$$\displaystyle{\|X(k)(\mathit{\mbox{ Id}} - P)X(-s)\| \leq 2r^{2}N^{2}\mu ^{s-k},\quad 0 \leq k \leq s.}$$

Proof

Fix a natural s and a unit vector ξ. Define a sequence y = {y k } by

$$\displaystyle{\,y_{k} = \left \{\begin{array}{ll} - X(k)(\mbox{ Id} - P)X(-s)\xi,&0 \leq k < s;\\ X(k)PX(-s)\xi, &k \geq s. \\ \end{array} \right.}$$

The sequence y coincides (up to a finite number of terms) with a solution of Eq. (2.2) with initial point from V 1; hence, \(y \in \mathcal{B}\).

Now we define a sequence f by

$$\displaystyle{\,f_{k} = \left \{\begin{array}{ll} 0,&k\neq s;\\ \xi, &k = s.\\ \end{array} \right.}$$

It is easily seen that the above sequence y is a solution of Eq. (2.3) with inhomogeneity f. Hence, y = T( f), and ∥ y∥ ≤ r.

The definition of y implies that

$$\displaystyle{\vert X(k)PX(-s)\xi \vert = \vert \,y_{k}\vert \leq r,\quad 0 \leq s \leq k.}$$

Since ξ is an arbitrary unit vector, ∥X(k)PX(−s)∥ ≤ r for 0 ≤ sk.

We replace ξ by the solution of the equation x s = X(s)ξ to show that

$$\displaystyle{ \vert X(k)P\xi \vert = \vert X(k)PX(-s)x_{s}\vert \leq r\vert x_{s}\vert,\quad 0 \leq s \leq k. }$$
(2.18)

Using inequalities (2.13), (2.16) with k 0 = s, and (2.18) with k = s, we see that

$$\displaystyle{\vert X(k)PX(-s)x_{s}\vert = \vert X(k)P\xi \vert \leq r\left (\sum _{u=s}^{k}\vert X(u)P\xi \vert ^{-1}\right )^{-1} \leq }$$
$$\displaystyle{\leq r\left (\mu ^{-(k-s)}\vert X(s)P\xi \vert ^{-1}\right )^{-1} = r\mu ^{k-s}\vert X(s)P\xi \vert \leq r^{2}\mu ^{k-s}\vert x_{ s}\vert.}$$

If = 0, then the resulting estimate is obvious. Since x s = X(s)ξ and X(s) is an isomorphism, we get the following estimate for the operator norm:

$$\displaystyle{\|X(k)PX(-s)\| \leq r^{2}\mu ^{k-s},\quad 1 \leq s \leq k.}$$

In this reasoning, we have used inequality (2.18) with s = k. It is also true for s = k = 0 since ∥ P∥ ≤ 1. Therefore, the first estimate of our lemma is proved for 0 ≤ sk.

The proof of the second estimate is quite similar. The only difference is as follows. We cannot use an analog of (2.18) with k = s since ks in the definition of the sequence y. The following inequality is proved by the same reasoning as above:

$$\displaystyle{\vert X(k)(\mbox{ Id} - P)\xi \vert = \vert X(k)(\mbox{ Id} - P)X(-s)x_{s}\vert \leq r\vert x_{s}\vert,\quad s > k.}$$

In the case k = s − 1, we write

$$\displaystyle{\vert X(s)(\mbox{ Id} - P)\xi \vert = \vert A_{s-1}X(s - 1)(\mbox{ Id} - P)X(-s)x_{s}\vert \leq }$$
$$\displaystyle{\leq \| A_{s-1}\|\vert X(s - 1)(\mbox{ Id} - P)X(-s)x_{s}\vert \leq rN\vert x_{s}\vert,}$$

and then repeat the reasoning of the first case. □

Lemma 2.1.4 shows that if we take constants C 0 = r 2 N and λ = μ and projections

$$\displaystyle{\,P_{k} = X(k)PX(-k)\quad \mbox{ and}\quad Q_{k} = X(k)(\mbox{ Id} - P)X(-k),}$$

then the operators F(k, l) generated by the sequence \(\mathcal{A}\) satisfy estimates (2.6) and (2.7) with C = C 0 and λ. Clearly, relations (2.4) and (2.5) are valid.

Thus, to show that \(\mathcal{A}\) is hyperbolic on I, it remains to prove the following statement.

Lemma 2.1.5

There exists a constant C = C(N, C 0, λ) ≥ C 0 such that inequalities ( 2.8 ) are fulfilled.

Proof

Let L 1 and L 2 be two linear subspaces of \(\mathbb{R}^{n}\). Introduce the value

$$\displaystyle{\angle (L_{1},L_{2}) =\min \vert v_{1} - v_{2}\vert,}$$

where the minimum is taken over all pairs of unit vectors v 1L 1, v 2L 2.

We claim that there exists a constant C 1 = C 1(N, C 0, λ) such that

$$\displaystyle{ \angle (S_{k},U_{k}) \geq C_{1},\quad k \in I. }$$
(2.19)

Fix an index kI, take unit vectors v 1S k and v 2U k for which (S k , U k ) = | v 1v 2 |, and denote

$$\displaystyle{\alpha _{l} = \vert \,F(l,k)(v_{1} - v_{2})\vert,\quad l \geq k.}$$

Inequalities (2.6) and (2.7) imply that

$$\displaystyle{\alpha _{l} \geq \vert \,F(l,k)v_{2}\vert -\vert \,F(l,k)v_{1}\vert \geq \lambda ^{k-l}/C_{ 0} - C_{0}\lambda ^{l-k}.}$$

Hence, there exists a constant m = m(C 0, λ) such that

$$\displaystyle{\alpha _{k+m} \geq 1.}$$

At the same time, it follows from (2.1) that

$$\displaystyle{\alpha _{k+m} \leq N^{m}\alpha _{ k}.}$$

Combining the above two inequalities, we see that

$$\displaystyle{\angle (S_{k},U_{k}) = \alpha _{k} \geq C_{1}(N,C_{0},\lambda ):= N^{-m(C_{0},\lambda )},}$$

which proves (2.19).

Clearly, if v 1 and v 2 are two unit vectors, then the usual angle 〈v 1, v 2〉 satisfies the relation

$$\displaystyle{\vert v_{1} - v_{2}\vert = 2\sin (\langle v_{1},v_{2}\rangle /2),}$$

and we see that estimate (2.19) implies the existence of β = β(N, C 0, λ) such that if γ is the usual angle between S k and U k , then

$$\displaystyle{\sin (\gamma ) \geq \beta.}$$

Now we take an arbitrary unit vector \(v \in \mathbb{R}^{n}\) and denote v s = P k v. If γ s is the angle between v and v s , then the sine law implies that

$$\displaystyle{\frac{\vert v\vert } {\sin (\gamma )} = \frac{\vert v_{s}\vert } {\sin (\gamma _{0})} \geq \vert v_{s}\vert,}$$

and we conclude that

$$\displaystyle{\vert v_{s}\vert = \vert \,P_{k}v\vert \leq 1/\beta,}$$

which implies that

$$\displaystyle{\|\,P_{k}\| \leq C =\max (C_{0},1/\beta ).}$$

A similar estimate holds for ∥Q k ∥. □

As we said above, the following statement holds.

Lemma 2.1.6

A hyperbolic sequence \(\mathcal{A}\) has the Perron property on I.

Proof

Assume that the sequence \(\mathcal{A}\) has properties stated in relations (2.4)–(2.8). Take a sequence

$$\displaystyle{\,f =\{\, f_{k} \in \mathbb{R}^{n}:\; k \in I\}}$$

such that ∥ f∥ = ν < and consider the sequence y defined by formula (2.11).

Then

$$\displaystyle{\vert X(k)PX(-u)f_{u}\vert \leq C\lambda ^{k-u}\nu,\quad 0 \leq u \leq k,}$$

and

$$\displaystyle{\vert X(k)(\mbox{ Id} - P)X(-u)f_{u}\vert \leq C\lambda ^{u-k}\nu,\quad k + 1 \leq u < \infty,}$$

which implies that the second term in (2.11) is a convergent series (hence, the sequence y is a solution of (2.3)) and the estimate

$$\displaystyle{\|\,y\| \leq C(1 + \lambda + \lambda ^{2}\ldots )\nu + C(\lambda + \lambda ^{2}\ldots )\nu = \frac{1 + \lambda } {1 -\lambda }C\nu }$$

holds. □

Now we pass to the Pliss theorem.

This time, \(I = \mathbb{Z}\), and we denote \(I_{+} =\{ k \in \mathbb{Z}:\; k \geq 0\}\) and \(I_{-} =\{ k \in \mathbb{Z}:\; k \leq 0\}\).

Now \(\mathcal{A}\) is a sequence of linear isomorphisms

$$\displaystyle{A_{k}:\; \mathbb{R}^{n} \rightarrow \mathbb{R}^{n},\quad k \in I = \mathbb{Z}.}$$

It is again assumed that an analog of inequalities (2.1) holds, and we consider difference equations (2.2) and (2.3).

The Perron property of (2.2) on \(\mathbb{Z}\) is defined literally as in the case of \(I =\{ k \in \mathbb{Z}:\; k \geq 0\}\).

It follows from the Maizel’ theorem and its obvious analog for the case of \(I =\{ k \in \mathbb{Z}:\; k \leq 0\}\) that the sequence \(\mathcal{A}\) is hyperbolic on both I + and I (the definition of hyperbolicity in the case of I is literally the same).

Without loss of generality, we assume that C and λ are the same for the hyperbolicity on I + and I and denote by S k +, U k +, kI +, and S k , U k , kI , the corresponding subspaces of \(\mathbb{R}^{n}\).

Theorem 2.1.2 (Pliss)

If \(\mathcal{A}\) has the Perron property on \(I = \mathbb{Z}\) , then the subspaces U 0 and S 0 + are transverse.

Remark 2.1.2

In fact, Pliss proved in [74] that the transversality of U 0 and S 0 + is equivalent to the Perron property of \(\mathcal{A}\) on \(I = \mathbb{Z}\), but we need only the implication stated above.

Remark 2.1.3

Note that there exist sequences \(\mathcal{A}\) that are separately hyperbolic on I + and I for which the subspaces U 0 and S 0 + are transverse and such that these sequences are not hyperbolic on \(I = \mathbb{Z}\). It is easy to construct such a sequence with \(S_{k}^{+} = \mathbb{R}^{n},U_{k}^{+} =\{ 0\},k \in I_{+},\) and \(S_{k}^{-} =\{ 0\},U_{k}^{-} = \mathbb{R}^{n},k \in I_{-}\) (we leave details to the reader).

Proof

To get a contradiction, assume that the subspaces U 0 and S 0 + are not transverse. Then there exists a vector \(x \in \mathbb{R}^{n}\) such that

$$\displaystyle{ x\neq y_{1} + y_{2} }$$
(2.20)

for any y 1U 0 and y 2S 0 +.

Since the subspaces U 0 + and S 0 + are complementary (see (2.4)), we can represent

$$\displaystyle{x =\xi +\eta,\quad \xi \in S_{0}^{+},\;\eta \in U_{ 0}^{+}.}$$

Then it follows from (2.20) that

$$\displaystyle{ \eta \neq z_{1} + z_{2} }$$
(2.21)

for any z 1S 0 + and z 2U 0 . We may assume that | η | = 1.

Consider the sequence

$$\displaystyle{a_{k} = \left \{\begin{array}{ll} 0,&k \leq 0;\\ 1, &k > 0.\\ \end{array} \right.}$$

Since η ≠ 0 in (2.21), X(k)η ≠ 0 for kI. Define a sequence f = { f k ,  kI} by

$$\displaystyle{ f_{k} = \frac{X(k)\eta } {\vert X(k)\eta \vert }a_{k},\quad k \in I. }$$
(2.22)

Clearly, ∥ f∥ = 1. We claim that the corresponding Eq. (2.3) does not have bounded solutions.

Consider the sequence

$$\displaystyle{\phi _{k} = -\sum _{u=k+1}^{\infty }X(k)(\mbox{ Id} - P)X(-u)f_{ u},\quad k \geq 0.}$$

In this formula, P is the projection defined for Eq. (2.2).

The sequence {ϕ k } is bounded for k ≥ 0. Indeed, f u U u + for u ≥ 0; hence,

$$\displaystyle{\vert \phi _{k}\vert = \left \vert \sum _{u=k+1}^{\infty }X(k)(\mbox{ Id} - P)X(-u)f_{ u}\right \vert \leq }$$
$$\displaystyle{\leq \sum _{u=k+1}^{\infty }C\lambda ^{u-k} = C \frac{\lambda } {1 -\lambda }.}$$

We know that since the series defining ϕ k is convergent, the sequence {ϕ k } is a solution of the homogeneous equation (2.2) for k ≥ 0.

Clearly,

$$\displaystyle{\phi _{0} = -\sum _{u=1}^{\infty }(\mbox{ Id} - P)X(-u)f_{ u} = -\sum _{u=1}^{\infty } \frac{\eta } {\vert X(u)\eta \vert } =\nu \eta,}$$

where

$$\displaystyle{\nu = -\sum _{u=1}^{\infty } \frac{1} {\vert X(u)\eta \vert }.}$$

Deriving these relations, we take into account the definition of f and the equality (Id − P)η = η. In addition, the value ν is finite since

$$\displaystyle{ \frac{1} {\vert X(k)\eta \vert } \leq C\lambda ^{k},\quad k \geq 0,}$$

due to inequalities (2.7).

It follows from (2.21) that

$$\displaystyle{ \phi _{0}\neq y_{1} + y_{2} }$$
(2.23)

for any y 1S 0 + and y 2U 0 .

Now let us assume that Eq. (2.2) has a solution ψ = { ψ k } that is bounded on \(I = \mathbb{Z}\). Then ψ 0U 0 .

On the other hand,

$$\displaystyle{\psi _{k} = X(k)(\psi _{0} -\phi _{0}) +\phi _{0}.}$$

Since ϕ k are bounded for k ≥ 0, ψ k can be bounded for k ≥ 0 only if

$$\displaystyle{X(k)(\psi _{0} -\phi _{0})}$$

are bounded for k ≥ 0, which implies that

$$\displaystyle{\psi _{0} -\phi _{0} \in S_{0}^{+}.}$$

Set

$$\displaystyle{\,y_{1} =\phi _{0} -\psi _{0} \in S_{0}^{+}\quad \mbox{ and}\quad y_{ 2} =\psi _{0} \in U_{0}^{-}.}$$

Then ϕ 0 = y 1 + y 2, and we get a contradiction with (2.23). □

Remark 2.1.4

We will apply the Maizel’ and Pliss theorems proved in this section in a slightly different situation.

We consider a diffeomorphism f of a smooth closed manifold M, fix a point xM and the trajectory \(\{x_{k} = f^{k}(x):\; k \in \mathbb{Z}\}\) of this point and define linear isomorphisms

$$\displaystyle{A_{k} = Df(x_{k}):\; T_{x_{k}}M \rightarrow T_{x_{k+1}}M.}$$

To the sequence \(\mathcal{A} =\{ A_{k}\}\) we assign difference equations

$$\displaystyle{v_{k+1} = A_{k}v_{k},\quad v_{k} \in T_{x_{k}}M,}$$

and

$$\displaystyle{v_{k+1} = A_{k}v_{k} + f_{k+1},\quad v_{k} \in T_{x_{k}}M,\;f_{k+1} \in T_{x_{k+1}}M.}$$

Clearly, these difference equations are completely similar to Eqs. (2.2) and (2.3), and analogs of the Maizel’ and Pliss theorems are valid for them.

Historical Remarks

Theorem 2.1.1 was proved by A. D. Maizel’ in [38]. See also the classical W. A. Coppel’s book [13].

The Pliss theorem (Theorem 2.1.2) was published in [74]. Later, it was generalized by many authors; let us mention, for example, K. Palmer [55] who studied Fredholm properties of the corresponding operators.

2.2 Mañé Theorem

In this section, we prove Theorem 1.3.7.

Remark 2.2.1

In several papers, the analytic strong transversality condition is formulated in the following form, which is obviously stronger than the condition formulated in Definition 1.3.11: it is assumed that

$$\displaystyle{\tilde{B}^{+}(x) +\tilde{ B}^{-}(x) = T_{ x}M,\quad x \in M,}$$

where the subspaces \(\tilde{B}^{+}(x)\) and \(\tilde{B}^{-}(x)\) are defined by the equalities

$$\displaystyle{\tilde{B}^{+}(x) = \left \{v \in T_{ x}M:\;\lim _{k\rightarrow \infty }\left \vert Df^{k}(x)v\right \vert = 0\right \}}$$

and

$$\displaystyle{\tilde{B}^{-}(x) = \left \{v \in T_{ x}M:\;\lim _{k\rightarrow -\infty }\left \vert Df^{k}(x)v\right \vert = 0\right \}.}$$

In fact, it is easily seen from our proof below that the structural stability of f implies this form of the analytic strong transversality condition as well, so that both conditions are equivalent.

The main part of our proof of Theorem 1.3.7 is contained in the following statement.

Theorem 2.2.1

The analytic strong transversality condition implies Axiom A.

First we prove that the analytic strong transversality condition implies the hyperbolicity of the nonwandering set Ω.

We assign to a diffeomorphism f: MM the mapping π: TMTM (where TM is the tangent bundle of M) which maps a pair (x, v) ∈ TM (where xM and vT x M) to the pair ( f(x), Df(x)v).

A subbundle Y of TM is a set of pairs (x, Y x ), where xM and Y x is a linear subspace of T x M.

Definition 2.2.1

A subbundle Y is called π-invariant if

$$\displaystyle{Df(x)Y _{x} = Y _{f(x)}\quad \mbox{ for}\quad x \in M.}$$

Assuming that f satisfies the analytic strong transversality condition, we define two subbundles B + and B of TM by setting

$$\displaystyle{B_{x}^{+} = B^{+}(x)\quad \mbox{ and}\quad B_{ x}^{-} = B^{-}(x)\quad \mbox{ for}\quad x \in M.}$$

Since

$$\displaystyle{\lim \inf _{k\rightarrow \infty }\left \vert Df^{k}(x)v\right \vert = 0}$$

if and only if

$$\displaystyle{\lim \inf _{k\rightarrow \infty }\left \vert Df^{k}(\,f(x))Df(x)v\right \vert = 0,}$$

the subbundle B + is π-invariant. A similar reasoning shows that the subbundle B is π-invariant as well.

The main object in the proof is the mapping π , dual to the mapping π.

Denote by <, > the scalar product in T x M. Let D f(x): T f(x) MT x M be defined as follows:

$$\displaystyle{<\xi,Df(x)v >=< D^{{\ast}}f(x)\xi,v >}$$

for all vT x M and ξT f(x) M (thus, D f(x) is the adjoint of Df(x)). We define π as follows: a pair \(\left (\,f(x),\xi \right ),\;\xi \in T_{f(x)}M,\) is mapped to

$$\displaystyle{\pi ^{{\ast}}\left (\,f(x),\xi \right ) = \left (x,D^{{\ast}}f(x)\xi \right ).}$$

If p: TMM is the projection to the first coordinate (i.e., p(x, v) = x), then p(π(x, v)) = f(x) (in this case, one says that π covers f); since p(π (x, v)) = f −1(x), π covers f −1.

Clearly, the definition of π implies the following statement.

Lemma 2.2.1

$$\displaystyle{(\pi ^{{\ast}})^{{\ast}} =\pi.}$$

If Y is a subbundle of TM, we define the orthogonal subbundle Y as follows:

$$\displaystyle{\,Y _{x}^{\perp } = \left \{\xi:\;<\xi,v >= 0\quad \mbox{ for all}\quad v \in Y _{ x}\right \},\quad x \in M.}$$

Lemma 2.2.2

If a subbundle Y is π-invariant, then Y is π -invariant.

Proof

Consider vectors ξY f(x) and D f(x)ξT x M. If vY x , then

$$\displaystyle{< v,D^{{\ast}}f(x)\xi >=<\xi,Df(x)v >= 0}$$

since Df(x)vY f(x), which means that D f(x)ξY x . □

We call two subbundles Y 1 and Y 2 complementary if

$$\displaystyle{ Y _{x}^{1} \oplus Y _{ x}^{2} = T_{ x}M\quad \mbox{ for any}\quad x \in M. }$$
(2.24)

Lemma 2.2.3

If Y 1 and Y 2 are complementary subbundles that are π-invariant, then \(\left (Y ^{1}\right )^{\perp }\) and \(\left (Y ^{2}\right )^{\perp }\) are complementary subbundles that are π -invariant.

Proof

The subbundles \(\left (Y ^{1}\right )^{\perp }\) and \(\left (Y ^{2}\right )^{\perp }\) are π -invariant by Lemma 2.2.2. If dimY x 1 = k, then equality (2.24) implies that dimY x 2 = nk. Clearly,

$$\displaystyle{ \mbox{ dim}\left (\,Y ^{1}\right )_{ x}^{\perp } = n - k\quad \mbox{ and }\quad \mbox{ dim}\left (\,Y ^{2}\right )_{ x}^{\perp } = k. }$$
(2.25)

Consider a vector \(\xi \in \left (Y ^{1}\right )_{x}^{\perp }\cap \left (Y ^{2}\right )_{x}^{\perp }\). Due to (2.24), any vector vT x M is representable as

$$\displaystyle{v = v_{1} + v_{2},\quad v_{1} \in Y _{x}^{1},\;v_{ 2} \in Y _{x}^{2}.}$$

Then < ξ, v > = < ξ, v 1 > + < ξ, v 2 > = 0. Since v is arbitrary, ξ = 0. The equality

$$\displaystyle{\left (\,Y ^{1}\right )_{ x}^{\perp }\cap \left (\,Y ^{2}\right )_{ x}^{\perp } =\{ 0\}}$$

and (2.25) imply the statement of our lemma. □

Let M 0M be a hyperbolic set of f. Then S and U defined by S x = S(x) and U x = U(x) for xM 0 are two complementary π-invariant subbundles on M 0 such that inequalities (HSD2.3) and (HSD2.4) hold (see Definition 1.3.1). In this case, we say that M 0 is hyperbolic with respect to π with subbundles S and U and constants C and λ.

Lemma 2.2.4

If a set M 0 is hyperbolic with respect to π with subbundles S and U and constants C and λ, then M 0 is hyperbolic with respect to π with subbundles U and S and the same constants C and λ.

Proof

If A and B are linear operators, then (AB) = B A ; hence,

$$\displaystyle{{\bigl (Df(\,f(x))Df(x)\bigr )}^{{\ast}} = D^{{\ast}}f(x)D^{{\ast}}f(\,f(x)).}$$

If we take vT x M and \(\xi \in T_{f^{2}(x)}M\), then

$$\displaystyle{< Df^{2}(x)v,\xi >=< Df(\,f(x))Df(x)v,\xi >=}$$
$$\displaystyle{=< Df(x)v,D^{{\ast}}f(\,f(x))\xi >=< v,D^{{\ast}}f(x)D^{{\ast}}f(\,f(x))\xi >=< v,D^{{\ast}}f^{2}(x)\xi >.}$$

Applying induction, it is easy to show that

$$\displaystyle{ < v,D^{{\ast}}f^{k}(x)\xi >=<\xi,Df^{k}(x)v >,\quad k \in \mathbb{Z}, }$$
(2.26)

for vT x M and \(\xi \in T_{f^{k}(x)}M\), where

$$\displaystyle{D^{{\ast}}f^{k}(x) = D^{{\ast}}f(x)D^{{\ast}}f(\,f(x))\ldots D^{{\ast}}f(\,f^{k-1}(x))}$$

and

$$\displaystyle{Df^{k}(x) = Df(\,f^{k-1}(x))Df(\,f^{k-2}(x))\ldots Df(x).}$$

By Lemma 2.2.3, the subbundles S and U are complementary and π -invariant.

Fix k ≥ 0 and a vector \(\xi \in \left (U_{f^{k}(x)}\right )^{\perp }\). Then D f k(x)ξT x M. The obvious equality

$$\displaystyle{\vert \eta \vert =\max _{\vert v\vert =1} <\eta,v >,\quad \eta,v \in T_{x}M,}$$

implies that

$$\displaystyle{\left \vert D^{{\ast}}f^{k}(x)\xi \right \vert =\max _{ \vert v\vert =1} < v,D^{{\ast}}f^{k}(x)\xi >.}$$

Represent v = v 1 + v 2, where v 1S x and v 2U x .

Since U is π -invariant,

$$\displaystyle{D^{{\ast}}f^{k}(x)\xi \in (U_{ x})^{\perp },}$$

and < v 2, D f k(x)ξ > = 0. It follows that

$$\displaystyle{\left \vert D^{{\ast}}f^{k}(x)\xi \right \vert =\max _{ \vert v_{1}\vert =1} < v_{1},D^{{\ast}}f^{k}(x)\xi >=\max _{ \vert v_{1}\vert =1} <\xi,Df^{k}(x)v_{ 1} >\leq C\lambda ^{k}\vert \xi \vert.}$$

In the last inequality, we used inequality (HSD2.3) and the obvious relation

$$\displaystyle{<\xi,v >\leq \vert \xi \vert \vert v\vert.}$$

A similar reasoning shows that

$$\displaystyle{\left \vert D^{{\ast}}f^{-k}(x)\xi \right \vert \leq C\lambda ^{-k}\vert \xi \vert }$$

for \(\xi \in \left (S_{f^{k}(x)}\right )^{\perp }\) and k ≤ 0. □

Now we prove that the analytic strong transversality condition implies that, in a sense, π does not have nontrivial bounded trajectories. Fix a point (x, v) ∈ TM and define the sequence (x k , v k ) = (π )k(x, v).

Lemma 2.2.5

If

$$\displaystyle{ \sup _{k\in \mathbb{Z}}\vert v_{k}\vert < \infty, }$$
(2.27)

then v = 0.

Proof

The obvious equalities

$$\displaystyle{x = f^{-k}\left (\,f^{k}(x)\right )\quad \mbox{ and}\quad u = Df^{-k}\left (\,f^{k}(x)\right )Df^{k}(x)u}$$

which are valid for all xM, uT x M, and \(k \in \mathbb{Z}\) imply that

$$\displaystyle{<\xi,u >=<\xi,Df^{-k}\left (\,f^{k}(x)\right )Df^{k}(x)u >=< D^{{\ast}}f^{-k}\left (\,f^{k}(x)\right )\xi,Df^{k}(x)u >}$$

for all ξ, uT x M and k.

Assume that a point (x, v) satisfies condition (2.27).

By the analytic strong transversality condition, we can represent any vector ξT x M in the form ξ = ξ 1 + ξ 2 for which there exist sequences l n and m n → − as n such that

$$\displaystyle{{\bigl |Df^{l_{n} }(x)\xi _{1}\bigr |} \rightarrow 0\quad \mbox{ and}\quad {\bigl |Df^{m_{n} }(x)\xi _{2}\bigr |} \rightarrow 0,\quad n \rightarrow \infty.}$$

Let us write

$$\displaystyle{< v,\xi >=< v,\xi _{1} +\xi _{2} >=< v,Df^{-l_{n} }\left (\,f^{l_{n} }(x)\right )Df^{l_{n} }(x)\xi _{1} > +}$$
$$\displaystyle{+ < v,Df^{-m_{n} }\left (\,f^{m_{n} }(x)\right )Df^{m_{n} }(x)\xi _{1} >=}$$
$$\displaystyle{ =< D^{{\ast}}f^{-l_{n} }\left (\,f^{l_{n} }(x)\right )v,Df^{l_{n} }(x)\xi _{1} > + < D^{{\ast}}f^{-m_{n} }\left (\,f^{m_{n} }(x)\right )v,Df^{m_{n} }(x)\xi _{2} >. }$$
(2.28)

By condition (2.27), both values \({\bigl |D^{{\ast}}f^{-l_{n}}(\,f^{l_{n}}(x))v\bigr |}\) and \({\bigl |D^{{\ast}}f^{-m_{n}}(\,f^{m_{n}}(x))v\bigr |}\) are bounded; hence, both terms in (2.28) tend to 0 as n. Thus, < ξ, v > = 0 for any ξ, which means that v = 0. □

To simplify notation, let us denote π by ρ and write

$$\displaystyle{\rho (x,v) = (\phi (x),\varPhi (x)v),}$$

so that ϕ(x) = f −1(x) and Φ(x) is the linear mapping T x MT ϕ(x) M, Φ(x) = D f(x). Let

$$\displaystyle{\,F(0,x) = \mbox{ Id},}$$
$$\displaystyle{\,F(k,x) =\varPhi (\phi ^{k-1}(x))\cdots \varPhi (x),\quad k > 0,}$$

and

$$\displaystyle{\,F(-k,x) =\varPhi ^{-1}(\phi ^{1-k}(x))\cdots \varPhi ^{-1}(x),\quad k > 0.}$$

Obviously, the mapping ρ is continuous. By Lemma 2.2.5, it satisfies the following Condition B: If

$$\displaystyle{\sup _{k\in \mathbb{Z}}\vert \,F(k,x)v\vert < \infty }$$

for some (x, v) ∈ TM, then v = 0.

Let us define the following two subbundles in TM: V = {(x, V x )} and W = {(x, W x )}. We agree that

  • vT x M belongs to V x if | F(k, x)v | → 0 as k

    and

  • vT x M belongs to W x if | F(k, x)v | → 0 as k → −.

Clearly, the subbundles V and W are ρ-invariant.

Lemma 2.2.6

Let a sequence (x m , v m ) ∈ TM be such that

  1. (1)

    (x m , v m ) → (x, v) as m∞;

  2. (2)

    there exists a number L > 0 and a sequence k m ∞ as m∞ such that

    $$\displaystyle{ \left \vert \,F(k,x_{m})v_{m}\right \vert \leq L,\quad 0 \leq k \leq k_{m}. }$$
    (2.29)

    Then (x, v) ∈ V.

Proof

Fix an arbitrary l ≥ 0. There exists an m 0 such that k m > l for mm 0. Then it follows from (2.29) that

$$\displaystyle{ \left \vert \,F(l,x_{m})v_{m}\right \vert \leq L. }$$
(2.30)

Since F(l, y)w is continuous in y and w, we may pass to the limit in (2.30) as m; thus,

$$\displaystyle{\vert \,F(l,x)v\vert \leq L.}$$

Since l is arbitrary, this means that

$$\displaystyle{ \vert \,F(k,x)v\vert \leq L,\quad k \geq 0. }$$
(2.31)

Let (x 0, v 0) be a limit point of the sequence \(\left (\phi ^{k}(x),F(k,x)v\right )\), i.e., the limit of the sequence

$$\displaystyle{ \left (\phi ^{t_{m} }(x),F(t_{m},x)v\right ) }$$
(2.32)

for some sequence t m .

Take an arbitrary \(k \in \mathbb{Z}\). Since

$$\displaystyle{\phi ^{t_{m} }(x) \rightarrow x_{0}\quad \mbox{ and}\quad F(t_{m},x)v \rightarrow v_{0},\quad m \rightarrow \infty,}$$
$$\displaystyle{ \phi ^{k+t_{m} }(x) \rightarrow \phi ^{k}(x_{ 0})\quad \mbox{ and}\quad F(k + t_{m},x)v \rightarrow F(k,x_{0})v_{0},\quad m \rightarrow \infty. }$$
(2.33)

For large m, k + t m > 0, and it follows from (2.31) and the second relation in (2.33) that

$$\displaystyle{ \left \vert \,F(k,x_{0})v_{0}\right \vert \leq L. }$$
(2.34)

Since (2.34) is valid for any \(k \in \mathbb{Z}\), Condition B implies that v 0 = 0. Thus, in any convergent sequence of the form (2.32) with t m ,

$$\displaystyle{\left \vert \,F(t_{m},x)v\right \vert \rightarrow 0,}$$

which means that (x, v) ∈ V. □

Remark 2.2.2

A similar reasoning shows that if we take k m → − and k m k ≤ 0 in condition (2) of Lemma 2.2.6, then (x, v) ∈ W. In what follows, we do not make such comments and only consider the case of the subbundle V.

Define the set

$$\displaystyle{A =\{ (x,v) \in TM:\; \vert \,F(k,x)v\vert \leq 1\quad \mbox{ for}\quad k \geq 1\}.}$$

Clearly, the set A is positively ρ-invariant, i.e., if (x, v) ∈ A and k ≥ 0, then \(\left (\phi ^{k}(x),F(k,x)v\right ) \in A\).

Let us say that a set C = {(x, v) ∈ TM} is bounded if

$$\displaystyle{\sup _{(x,v)\in C}\vert v\vert < \infty.}$$

Since the manifold M is compact, any closed and bounded subset C of TM is (sequentially) compact, i.e., any sequence in C has a convergent subsequence, and the limit of this subsequence belongs to C.

Lemma 2.2.7

The set A is a compact subset of V.

Proof

It was shown in the proof of Lemma 2.2.6 that inequality (2.31) implies the inclusion (x, v) ∈ V; thus, AV. Since F(0, x)v = v, A is bounded. Consider a sequence (x m , v m ) ∈ A such that (x m , v m ) → (x, v),  m. For any fixed k ≥ 0,

$$\displaystyle{\vert \,F(k,x)v\vert =\lim _{m\rightarrow \infty }\vert \,F(k,x_{m})v_{m}\vert \leq 1.}$$

Hence, (x, v) ∈ A, and A is closed. □

Lemma 2.2.8

For any μ > 0 there exists a K > 0 such that if (x, v) ∈ A, then

$$\displaystyle{ \vert \,F(k,x)v\vert <\mu,\quad k \geq K. }$$
(2.35)

Proof

Assuming the converse, let us find sequences (x m , v m ) ∈ A and k m and a number μ > 0 such that

$$\displaystyle{ \vert \,F(k_{m},x_{m})v\vert \geq \mu. }$$
(2.36)

Since A is positively ρ-invariant,

$$\displaystyle{\left (\phi ^{k_{m} }(x_{m}),F(k_{m},x_{m})v_{m}\right ) \in A;}$$

since A is compact, the above sequence has a convergent subsequence. Assume, for definiteness, that

$$\displaystyle{\left (\phi ^{k_{m} }(x_{m}),F(k_{m},x_{m})v_{m}\right ) \rightarrow (x,v).}$$

Then it follows from (2.36) that | v | ≥ μ. Fix a number \(k \in \mathbb{Z}\). Since k + k m > 0 for large m,

$$\displaystyle{\left (\phi ^{k+k_{m} }(x_{m}),F(k + k_{m},x_{m})v_{m}\right ) \rightarrow \left (\phi ^{k}(x),F(k,x)v\right ),\quad m \rightarrow \infty,}$$

and

$$\displaystyle{\left \vert \,F(k + k_{m},x_{m})v_{m})\right \vert \leq 1,}$$

we conclude that

$$\displaystyle{\vert \,F(k,x)v)\vert \leq 1,\quad k \in \mathbb{Z}.}$$

Condition B implies that v = 0. The contradiction with (2.36) completes the proof. □

Lemma 2.2.9

There exists a number μ > 0 such that if (x, v) ∈ V and | v | ≤ μ, then (x, v) ∈ A.

Proof

Assuming the contrary, we can find a sequence (x m , v m ) ∈ V such that | v m | → 0,  m, and (x m , v m ) ∉ A.

Then

$$\displaystyle{\mu _{m} =\max _{k\geq 0}\left \vert \,F(k,x_{m})v_{m}\right \vert > 1}$$

(we take into account that | F(k, x m )v m | → 0,  k).

Find numbers k m > 0 such that

$$\displaystyle{\left \vert \,F(k_{m},x_{m})v_{m}\right \vert =\mu _{m}.}$$

Since

$$\displaystyle{\vert \,F(k,x_{m})(v_{m}/\mu _{m})\vert \leq 1,\quad k \geq 0,}$$

(x m , v m μ m ) ∈ A.

The mapping ρ is continuous and F(k, x)0 = 0; hence,

$$\displaystyle{\max _{0\leq k\leq K}\vert \,F(k,x_{m})(v_{m}/\mu _{m})\vert \rightarrow 0,\quad m \rightarrow \infty,}$$

for any fixed K (note that x m M, M is compact, | v m | → 0, and μ m > 1).

Hence, k m ,  m. Lemma 2.2.8 implies now that the relations

$$\displaystyle{(x_{m},v_{m}/\mu _{m}) \in A\quad \mbox{ and}\quad \vert \,F(k_{m},x_{m})(v_{m}/\mu _{m})\vert = 1}$$

are contradictory. □

Lemma 2.2.10

There exists a number K > 0 such that if (x, v) ∈ V, then

$$\displaystyle{ \vert \,F(k,x)v\vert \leq (1/2)\vert v\vert,\quad k \geq K. }$$
(2.37)

Proof

Apply Lemma 2.2.8 to find a number K such that

$$\displaystyle{\vert \,F(k,x)v'\vert <\mu /2,\quad k \geq K,}$$

for any (x, v′) ∈ A (where μ is the number from Lemma 2.2.9).

Take any (x, v) ∈ V. If v ≠ 0, set v′ = μ(v∕ | v | ). Then (x, v′) ∈ A by Lemma 2.2.9, and it follows from Lemma 2.2.8 that

$$\displaystyle{\left \vert \,F(k,x)v'\right \vert = (\mu /\vert v\vert )\left \vert \,F(k,x)v\right \vert \leq \mu /2,\quad k \geq K,}$$

which obviously implies the desired relation (2.37). If v = 0, we have nothing to prove. □

Lemma 2.2.11

  1. (1)

    The subbundles V and W are closed.

  2. (2)

    There exist numbers C > 0 and λ ∈ (0, 1) such that

    if (x, v) ∈ V, then

    $$\displaystyle{ \vert \,F(k,x)v\vert \leq C\lambda ^{k}\vert v\vert,\quad k \geq 0; }$$
    (2.38)

    if (x, v) ∈ W, then

    $$\displaystyle{ \vert \,F(k,x)v\vert \leq C\lambda ^{-k}\vert v\vert,\quad k \leq 0. }$$
    (2.39)

Proof

We prove the statements for the subbundle V; for W, the proofs are similar.

To prove statement (1), consider a sequence (x k , v k ) ∈ V such that (x k , v k ) → (x, v) as k.

If v = 0, then, obviously, (x, v) ∈ V. Assume that v ≠ 0; then v k ≠ 0 for large k, and, by Lemma 2.2.9 there exists a μ > 0 such that

$$\displaystyle{(x_{k},\mu v_{k}/\vert v_{k}\vert ) \in A.}$$

Since A is closed (see Lemma 2.2.7),

$$\displaystyle{(x,\mu v/\vert v\vert ) \in A,}$$

and (x, v) ∈ V by Lemma 2.2.7. This proves the first statement of our lemma.

To prove the second one, apply Lemma 2.2.10 and find a number K such that

$$\displaystyle{ \vert \,F(k,x)v\vert \leq (1/2)\vert v\vert,\quad k \geq K, }$$
(2.40)

for any (x, v) ∈ V.

It follows from (2.40) and from the ρ-invariance of V that

$$\displaystyle{ \vert \,F(2K,x)v\vert \leq (1/2)^{2}\vert v\vert,\ldots,\vert \,F(kK,x)v\vert \leq (1/2)^{k}\vert v\vert,\quad k \geq 0. }$$
(2.41)

There exists a number C 0 > 0 such that

$$\displaystyle{ \max _{0\leq k<K,\;x\in M}\|\,F(k,x)\| \leq C_{0}. }$$
(2.42)

Let us show that inequality (2.38) holds with C = 2C 0 and λ = 21∕K. We can represent any k ≥ 0 in the form k = k 0 K + k 1, where k 0 ≥ 0 and 0 ≤ k 1 < K. If (x, v) ∈ V, then it follows from (2.41) and (2.42) that

$$\displaystyle{\vert \,F(k,x)v\vert = \vert \,F(k_{1},\phi ^{k_{0}K}(x))F(k_{ 0}K,x)v\vert \leq C_{0}(1/2)^{k_{0} }\vert v\vert,}$$

but since k 0 + 1 > kK, − k 0 < −kK + 1, and \(2^{-k_{0}} < 2\lambda ^{k}\), we conclude that

$$\displaystyle{\vert \,F(k,x)v\vert \leq C\lambda ^{k}\vert v\vert,}$$

as required. □

Remark 2.2.3

Inequalities (2.38) and (2.39) have the same form as inequalities (HSD2.3) and (HSD2.4) in the definition of a hyperbolic set. Thus, if we want to show that some compact, ρ-invariant subset M 0 of M is a hyperbolic set of ρ with subbundles V and W, we only have to show that

$$\displaystyle{ V _{x} + W_{x} = T_{x}M,\quad x \in M_{0}. }$$
(2.43)

Lemma 2.2.12

Assume that for a sequence (x m , v m ) ∈ TM there exists a sequence k m ∞ as m∞ and a number r > 0 such that

$$\displaystyle{\vert v_{m}\vert \leq r\quad \mathit{\mbox{ and}}\quad \vert \,F(k_{m},x_{m})v_{m}\vert \leq r.}$$

Then there exists a number R > 0 such that

$$\displaystyle{\left \vert \,F(k,x_{m})v_{m}\right \vert \leq R,\quad 0 \leq k \leq k_{m}.}$$

Proof

Assume the contrary, and let there exist (x m , v m ) ∈ TM and k m such that

$$\displaystyle{b_{m}:=\max _{0\leq k\leq k_{m}}\left \vert \,F(k,x_{m})v_{m}\right \vert \rightarrow \infty,\quad m \rightarrow \infty.}$$

Find numbers l m ∈ [0, k m ] such that \(b_{m} = \left \vert \,F(l_{m},x_{m})v_{m}\right \vert \). Since ρ is continuous, it is obvious that

$$\displaystyle{ l_{m} \rightarrow \infty \quad \mbox{ and}\quad k_{m} - l_{m} \rightarrow \infty,\quad m \rightarrow \infty. }$$
(2.44)

Set

$$\displaystyle{w_{m} = F(l_{m},x_{m})(v_{m}/b_{m}).}$$

Let (x, v) be a limit point of the sequence \((\phi ^{l_{m}}(x_{m}),w_{m})\); then | v | = 1. The inequality

$$\displaystyle{\left \vert \,F(k,\phi ^{l_{m} }(x_{m}))w_{m}\right \vert \leq 1}$$

holds for k ∈ [−l m , 0] ∪ [0, k m l m ]. We apply relations (2.44) and Lemma 2.2.6 (and its analog for W) to conclude that vV x W x , but then v = 0 by Condition B. □

Remark 2.2.4

A similar statement is valid if k m → −. In this case,

$$\displaystyle{\left \vert \,F(k,x_{m})v_{m}\right \vert \leq R,\quad k_{m} \leq k \leq 0.}$$

Lemma 2.2.13

If x is a nonwandering point of the diffeomorphism f, then equality ( 2.43 ) holds.

Proof

By the definition of a nonwandering point, there exist sequences of points x m M and numbers k m such that

$$\displaystyle{x_{m} \rightarrow x,\quad f^{k_{m} }(x_{m}) \rightarrow x,\quad \vert k_{m}\vert \rightarrow \infty }$$

as m. We may assume that k m → −.

Consider the linear subspace W x and let Q be its orthogonal complement. Let dimQ = s. Fix an orthonormal base v 1, , v s in Q. Clearly, we can find s orthonormal vectors v 1 m, , v s m in \(T_{x_{m}}M\) such that v j mv j as m for j = 1, , s.

Let Q m be the subspace of \(T_{x_{m}}M\) spanned by v 1 m, , v s m. Introduce the numbers

$$\displaystyle{\mu _{m} =\min \left \{\left \vert \,F(k_{m},x_{m})v\right \vert:\; v \in Q_{m},\;\vert v\vert = 1\right \}.}$$

We claim that

$$\displaystyle{ \mu _{m} \rightarrow \infty,\quad m \rightarrow \infty. }$$
(2.45)

If we assume the contrary, we can find a number r > 0 and sequences w m Q m , | w m | = 1, and k m → − such that

$$\displaystyle{\vert \,F(k_{m},x_{m})w_{m}\vert \leq r.}$$

By the remark to Lemma 2.2.12, there exists a number R such that

$$\displaystyle{\vert \,F(k,x_{m})w_{m}\vert \leq R,\quad k \in [k_{m},0].}$$

By Lemma 2.2.6, in this case, any limit point (x, v) of the sequence (x m , w m ) belongs to W, i.e., vW x . This relation contradicts our construction since w m Q m , which implies that v is orthogonal to Q (note that | v | = 1). This proves (2.45).

Consider the linear space

$$\displaystyle{K_{m} = F(k_{m},x_{m})Q_{m}.}$$

Clearly, \(K_{m} \subset T_{y_{m}}M\), where \(y_{m} = f^{k_{m}}(x_{m})\), and dimK m = s.

Consider a vector wK m , | w | = 1. Let w = F(k m , x m )v. It follows from the definition of the numbers μ m that

$$\displaystyle{ \vert v\vert \leq \mu _{m}\vert w\vert =\mu _{m}. }$$
(2.46)

Inequalities (2.46), relations (2.45), and Lemma 2.2.12 imply that for any sequence ( y m , w m ), where w m K m and | w m | = 1, there exists a number R such that

$$\displaystyle{\vert \,F(k,x_{m})w_{m}\vert \leq R,\quad k \in [0,-k_{m}].}$$

Now Lemma 2.2.6 implies that any limit point (x, w) of such a sequence ( y m , w m ) belongs to V, i.e., wV x .

Select an orthonormal basis w 1 m, , w s m in K m . We may assume that all the sequences w 1 m, , w s m converge for some sequence of indices. For definiteness, let

$$\displaystyle{w_{1}^{m} \rightarrow w_{ 1},\ldots,w_{s}^{m} \rightarrow w_{ s},\quad m \rightarrow \infty.}$$

The vectors w 1, , w s are pairwise orthogonal unit vectors in V x ; hence,

$$\displaystyle{ \mbox{ dim}V _{x} \geq s. }$$
(2.47)

By the definition of the spaces Q and Q m ,

$$\displaystyle{\mbox{ dim}W_{x} = n - s.}$$

Combining this with inequality (2.47), we see that

$$\displaystyle{\mbox{ dim}V _{x} + \mbox{ dim}W_{x} \geq n.}$$

Since V x W x = {0} by Condition B, we conclude that

$$\displaystyle{V _{x} + W_{x} = T_{x}M,}$$

as claimed. □

The nonwandering set of the diffeomorphism f coincides with the nonwandering set of the diffeomorphism ϕ = f −1. Combining Lemma 2.2.1 with Lemma 2.2.4 applied to the mapping ρ, we conclude that the following statement holds.

Theorem 2.2.2

If a diffeomorphism f satisfies the analytic strong transversality condition, then the nonwandering set of f is hyperbolic.

Now we show that the analytic strong transversality condition implies the second part of Axiom A, the density of periodic points in the nonwandering set Ω( f) of the diffeomorphism f.

Since we are going to use the Mañé theorem in the proof of the implication (the analytic strong transversality condition) ⇒ (structural stability) for a diffeomorphism f having the Lipschitz shadowing property, we can essentially simplify this proof (compared to the original Mañé proof) assuming that f has the shadowing property.

Thus, now we prove the following statement.

Theorem 2.2.3

If a diffeomorphism f has the shadowing property and the nonwandering set Ω( f) of f is hyperbolic, then periodic points are dense in Ω( f).

In this proof, we apply the following two well-known results (see, for example, [71] for their proofs).

First we recall a known definition.

Definition 2.2.2

A homeomorphism f of a metric space (M, dist) is called expansive on a set A with expansivity constant a > 0 if the relations

$$\displaystyle{\,f^{k}(x),f^{k}(\,y) \in A,\quad k \in \mathbb{Z},}$$

and

$$\displaystyle{\mbox{ dist}\left (\,f^{k}(x),f^{k}(\,y)\right ) \leq a,\quad k \in \mathbb{Z},}$$

imply that x = y.

Theorem 2.2.4

If Λ is a hyperbolic set of a diffeomorphism f, then there exists a neighborhood of Λ on which f is expansive.

Denote by cardA the cardinality of a finite or countable set A.

Theorem 2.2.5 (The Birkhoff Constant Theorem)

If the phase space X of a homeomorphism f is compact and U is a neighborhood of the nonwandering set Ω( f) of f, then there exists a constant T = T(U) such that for any point xX, the inequality

$$\displaystyle{\mathit{\mbox{ card}}\left \{k \in \mathbb{Z}:\; f^{k}(x)\notin U\right \} \leq T}$$

holds.

Proof (of Theorem 2.2.3)

Fix an arbitrary point zΩ( f). There exist sequences of points z n and numbers l n such that

$$\displaystyle{z_{n} \rightarrow z\quad \mbox{ and}\quad f^{l_{n} }(z_{n}) \rightarrow z,\quad n \rightarrow \infty.}$$

Let U be a neighborhood of the set Ω( f) on which f is expansive and let a be the corresponding expansivity constant.

Fix an ɛ > 0 such that the 3ɛ-neighborhood of Ω( f) is a subset of U. Denote by U′ the 2ɛ-neighborhood of Ω( f). We assume, in addition, that 2ɛ < a.

For this ɛ there exists a d > 0 such that any d-pseudotrajectory of f is ɛ-shadowed by an exact trajectory.

Fix an index n such that

$$\displaystyle{\mbox{ dist}(z,z_{n}),\,\mbox{ dist}(z,f^{l_{n} }(z_{n})) < d/2.}$$

Construct a sequence {x k } as follows. Represent \(k \in \mathbb{Z}\) in the form k = k 0 + k 1 l n , where \(k_{1} \in \mathbb{Z}\) and 0 ≤ k 0 < l n , and set \(x_{k} = f^{k_{0}}(z_{n})\).

Clearly, the sequence {x k } is periodic with period l n ; the choice of n implies that this sequence is a d-pseudotrajectory of f.

We claim that

$$\displaystyle{ \{x_{k}\} \subset U'. }$$
(2.48)

Assuming the contrary, we can find an index m such that x m U′, i.e.,

$$\displaystyle{\mbox{ dist}\left (x_{m},\varOmega (\,f)\right ) \geq 2\varepsilon,}$$

but then

$$\displaystyle{ \mbox{ dist}\left (x_{m+kl_{n}},\varOmega (\,f)\right ) \geq 2\varepsilon,\quad k \in \mathbb{Z}. }$$
(2.49)

Let pM be a point whose trajectory ɛ-shadows {x k }, i.e.,

$$\displaystyle{\mbox{ dist}\left (\,f^{k}(\,p),x_{ k}\right ) <\varepsilon,\quad k \in \mathbb{Z};}$$

let p k = f k( p).

Then it follows from inequalities (2.49) that

$$\displaystyle{\mbox{ dist}\left (\,p_{m+kl_{n}},\varOmega (\,f)\right ) \geq \varepsilon,\quad k \in \mathbb{Z},}$$

which contradicts Theorem 2.2.5. Thus, we have established inclusion (2.48).

Set \(r = f^{l_{n}}(\,p)\). Since \(x_{k} = x_{k+l_{n}}\), the following inequalities hold:

$$\displaystyle{\mbox{ dist}\left (\,f^{k}(r),x_{ k}\right ) = \mbox{ dist}\left (\,f^{k+l_{n} }(\,p),x_{k+l_{n}}\right ) <\varepsilon,\quad k \in \mathbb{Z}.}$$

Then

$$\displaystyle{\mbox{ dist}\left (\,f^{k}(r),f^{k}(\,p)\right ) < 2\varepsilon < a,\quad k \in \mathbb{Z};}$$

in addition, inclusion (2.48) implies that

$$\displaystyle{\,f^{k}(r),f^{k}(\,p) \in U,\quad k \in \mathbb{Z}.}$$

Since f is expansive on U, r = p.

Thus, p is a periodic point of f.

Since ɛ and d can be taken arbitrarily small, there is such a point p in an arbitrarily small neighborhood of the point z. □

Thus, it remains to show that the analytic strong transversality condition implies the strong transversality condition (stable and unstable manifolds of nonwandering points are transverse).

For this purpose, we apply the following well-known theorem on the behavior of trajectories of a diffeomorphism in a neighborhood of a hyperbolic set (its proof can be easily reduced to Theorem 6.4.9 in the book [28]).

Theorem 2.2.6

Let Λ be a hyperbolic set of a diffeomorphism f with hyperbolicity constants C, λ. For any C 1 > C and λ 1 ∈ (λ, 1) there exists a neighborhood U of Λ with the following property. If xW s( p), pΛ, and f k(x) ∈ U for k ≥ 0, then there exist two complementary linear subspaces L +(x) and L (x) of T x M such that

  1. (1)
    $$\displaystyle{L^{+}(x) = T_{ x}W^{s}(\,p),\;L^{-}(x) = T_{ x}W^{u}(\,p);}$$
  2. (2)
    $$\displaystyle{\left \vert Df^{k}(x)v\right \vert \leq C_{ 1}\lambda _{1}^{k}\vert v\vert,\quad k \geq 0,\;v \in L^{+}(x),}$$

    and

    $$\displaystyle{\left \vert Df^{k}(x)v\right \vert \geq (1/C_{ 1})\lambda _{1}^{-k}\vert v\vert,\quad k \geq 0,\;v \in L^{-}(x).}$$

Remark 2.2.5

Of course, a similar statement holds if xW u( p), pΛ, and f k(x) belongs to a small neighborhood of Λ for k ≤ 0.

Clearly, it is enough for us to prove that if rW s( p) ∩ W u(q), where p, qΩ( f), then

$$\displaystyle{ B^{+}(r) \subset T_{ r}W^{s}(\,p)\mbox{ and }B^{-}(r) \subset T_{ r}W^{u}(q). }$$
(2.50)

We prove the first inclusion in (2.50) by proving that

$$\displaystyle{ B^{+}(r) \subset L^{+}(r) }$$
(2.51)

and applying Theorem 2.2.6; the second inclusion is proved in a similar way.

Any trajectory of a diffeomorphism satisfying Axiom A tends to one of the basic sets as time tends to ± (see Theorem 1.3.2).

Take as Λ the basic set to which f k(r) tends as k; obviously, p belongs to this basic set. Of course, we may assume that the positive semitrajectory of r belongs to a neighborhood of Λ having the properties described in Theorem 2.2.6.

Assume that inclusion (2.51) does not hold; take vB +(r)∖L +(r) and represent

$$\displaystyle{v = v^{s} + v^{u},\quad v^{s} \in L^{+}(r),\;v^{u} \in L^{-}(r);}$$

then v u ≠ 0.

Then

$$\displaystyle{\left \vert Df^{k}v\right \vert \geq \left \vert Df^{k}v_{ u}\right \vert -\left \vert Df^{k}v_{ s}\right \vert \geq (1/C_{1})\lambda ^{-k}\vert v^{u}\vert - C_{ 1}\lambda ^{k}\vert v^{s}\vert \rightarrow \infty,\quad k \rightarrow \infty,}$$

which contradicts the relation defining B +(r).

We have completely proved the Mañé theorem.

Historical Remarks

In his paper [39], R. Mañé gave several equivalent characterizations of structural stability of a diffeomorphism; Theorem 1.3.7 of this book is just one of them.

The property of expansivity of a dynamical system with discrete time is now one of the classical properties studied in the global theory of dynamical systems. Theorem 2.2.4 is folklore. Let us mention J. Ombach’s paper [49] in which it was shown (see Proposition 9) that a compact invariant set Λ of a diffeomorphism f is hyperbolic if and only if f | Λ is expansive and has the (standard) shadowing property (compare with Sect. 4.1).

Theorem 2.2.5 was proved in G. Birkhoff’s book [10].

2.3 Diffeomorphisms with Lipschitz Shadowing

Our main result in this section is as follows.

Theorem 2.3.1

If a diffeomorphism of class C 1 of a smooth closed n-dimensional manifold M has the Lipschitz shadowing property, then f is structurally stable.

As stated in Theorem 1.4.1 (1), a structurally stable diffeomorphism f has the Lipschitz shadowing property. Combining this statement with Theorem 2.3.1, we conclude that for diffeomorphisms, structural stability is equivalent to Lipschitz shadowing.

Proof (of Theorem 2.3.1)

Let us first explain the main idea of the proof.

Fix an arbitrary point pM, consider its trajectory \(\{\,p_{k} = f^{k}(\,p):\; k \in \mathbb{Z}\}\), and denote A k = Df( p k ). Consider the sequence \(\mathcal{A} =\{ A_{k}:\; k \in \mathbb{Z}\}\).

In Sect. 2.1 devoted to the Maizel’ and Pliss theorems, we worked with sequences \(\mathcal{A}\) of isomorphisms of Euclidean spaces. Here we apply these theorems (and all the corresponding notions of the Perron property etc.) to the sequences \(\mathcal{A} =\{ Df(\,p_{k})\}\) (see the remark concluding Sect. 3.1).

We claim that if f has the Lipschitz shadowing property, then \(\mathcal{A}\) has the Perron property on \(\mathbb{Z}\).

By the Maizel’ theorem, the Perron property on \(\mathbb{Z}\) implies that the sequence \(\mathcal{A}\) is hyperbolic on both “rays” \(\mathbb{Z}_{-}\) and \(\mathbb{Z}_{+}\). Denote by \(S_{k}^{-},U_{k}^{-},k \in \mathbb{Z}_{-}\) and \(S_{k}^{+},U_{k}^{+},k \in \mathbb{Z}_{+}\) the corresponding stable and unstable subspaces.

Then, by the Pliss theorem, the subspaces U 0 and S 0 + are transverse.

Clearly,

$$\displaystyle{\left \vert A_{k} \circ \ldots \circ A_{0}v\right \vert \rightarrow 0,\quad v \in S_{0}^{+},k \rightarrow \infty,}$$

and

$$\displaystyle{\left \vert (A_{k})^{-1} \circ \ldots \circ (A_{ 0})^{-1}v\right \vert \rightarrow 0,\quad v \in U_{ 0}^{-},k \rightarrow -\infty,}$$

which means that U 0 B ( p) and S 0 +B +( p), where B ( p) and B +( p) are the subspaces from the analytic transversality condition.

The transversality of the subspaces U 0 and S 0 + implies the transversality of the subspaces B (x) and B +(x). Since x is arbitrary, f is structurally stable by the Mañé theorem.

Now we prove our claim.

To clarify the reasoning, we first prove an analog of this result, Lemma 2.3.2, for a diffeomorphism of the Euclidean space \(\mathbb{R}^{n}\). Of course, \(\mathbb{R}^{n}\) is not compact, but we avoid the appearing difficulty making the following additional assumption (and noting that an analog of this assumption is certainly valid for a diffeomorphism of class C 1 of a closed smooth manifold). We call the condition below Condition S. Thus, we assume that for any μ > 0 we can find a δ = δ(μ) > 0 (independent of k) such that if | v | ≤ δ, then

$$\displaystyle{ \left \vert \,f(\,p_{k} + v) - A_{k}v - p_{k+1}\right \vert \leq \mu \vert v\vert,\quad k \in \mathbb{Z}. }$$
(2.52)

The basic technical part of the proof of Lemma 2.3.2 is the following statement (Lemma 2.3.1). In the following two Lemmas, 2.3.1 and 2.3.2, f is a diffeomorphism of \(\mathbb{R}^{n}\) that has the Lipschitz shadowing property with constants \(\mathcal{L},d_{0} > 0\), { p k = f k( p)} is an arbitrary trajectory of f, A k = Df( p k ), and it is assumed that Condition S is satisfied.

Lemma 2.3.1

Fix a natural number N. For any sequence

$$\displaystyle{w_{k} \in \mathbb{R}^{n},\quad k \in \mathbb{Z},}$$

with | w k | < 1 there exists a sequence

$$\displaystyle{z_{k} \in \mathbb{R}^{n},\quad k \in \mathbb{Z},}$$

such that

$$\displaystyle{ \vert z_{k}\vert \leq \mathcal{L} + 1,\quad k \in \mathbb{Z}, }$$
(2.53)

and

$$\displaystyle{ z_{k+1} = A_{k}z_{k} + w_{k+1},\quad - N \leq k \leq N. }$$
(2.54)

Proof

Thus, we assume that f has the Lipschitz shadowing property with constants \(\mathcal{L},d_{0} > 0\).

Define vectors

$$\displaystyle{\varDelta _{k} \in \mathbb{R}^{n},\quad - N \leq k \leq N + 1,}$$

by the following relations:

$$\displaystyle{ \varDelta _{-N} = 0\quad \mbox{ and}\quad \varDelta _{k+1} = A_{k}\varDelta _{k} + w_{k+1},\quad - N \leq k \leq N. }$$
(2.55)

Clearly, there exists a number Q (depending on N, \(\mathcal{A}\), and w k ) such that

$$\displaystyle{ \vert \varDelta _{k}\vert \leq Q,\quad - N \leq k \leq N + 1. }$$
(2.56)

Fix a small number d ∈ (0, d 0) (we will reduce this number during the proof) and consider the following sequence \(\xi =\{ x_{k} \in \mathbb{R}^{n}:\; k \in \mathbb{Z}\}\):

$$\displaystyle{x_{k} = \left \{\begin{array}{ll} f^{k+N}(\,p_{-N}), &k < -N; \\ p_{k} + d\varDelta _{k}, & - N \leq k \leq N + 1; \\ f^{k-N-1}(\,p_{N+1} + d\varDelta _{N+1}),&k > N + 1.\\ \end{array} \right.}$$

Note that if − NkN, then

$$\displaystyle{\left \vert x_{k+1} - f(x_{k})\right \vert = \left \vert \,p_{k+1} + d\varDelta _{k+1} - f(\,p_{k} + d\varDelta _{k})\right \vert \leq }$$
$$\displaystyle{\leq d\left \vert \varDelta _{k+1} - A_{k}\varDelta _{k}\right \vert + \left \vert \,f(\,p_{k} + d\varDelta _{k}) - p_{k+1} - dA_{k}\varDelta _{k}\right \vert.}$$

Since we consider a finite number of w k , the condition | w k | < 1 implies that there is a μ ∈ (0, 1) such that the first term above does not exceed μd; by Condition S, the second term is less than (1 −μ)d if d is small. Hence, in this case, the sum is less than d.

For the remaining values of k,

$$\displaystyle{\vert x_{k+1} - f(x_{k})\vert = 0.}$$

Thus, we may take dd 0 so small that ξ is a d-pseudotrajectory of f. Then there exists a trajectory \(\eta =\{ y_{k}:\; k \in \mathbb{Z}\}\) of f such that

$$\displaystyle{ \vert x_{k} - y_{k}\vert \leq \mathcal{L}d,\quad k \in \mathbb{Z}. }$$
(2.57)

Denote t k = ( y k p k )∕d. Since Δ k = (x k p k )∕d, it follows from (2.57) that

$$\displaystyle{ \vert \varDelta _{k} - t_{k}\vert = \vert x_{k} - y_{k}\vert /d \leq \mathcal{L},\quad k \in \mathbb{Z}. }$$
(2.58)

It follows from (2.56) and (2.57) that

$$\displaystyle{\vert \,y_{k} - p_{k}\vert \leq \vert \,y_{k} - x_{k}\vert + \vert x_{k} - p_{k}\vert \leq (\mathcal{L} + Q)d,\quad k \in \mathbb{Z}.}$$

Hence,

$$\displaystyle{ \vert t_{k}\vert \leq \mathcal{L} + Q,\quad k \in \mathbb{Z}. }$$
(2.59)

Now we define a finite sequence

$$\displaystyle{b_{k} \in \mathbb{R}^{n},\quad - N \leq k \leq N + 1,}$$

by the following relations:

$$\displaystyle{ b_{-N} = t_{-N}\quad \mbox{ and}\quad b_{k+1} = A_{k}b_{k},\quad - N \leq k \leq N. }$$
(2.60)

Take μ 1 ∈ (0, 1) such that

$$\displaystyle{ \left ((K + 1)^{2N} + (K + 1)^{2N-1} +\ldots +1\right )\mu _{ 1} < 1, }$$
(2.61)

where K = sup∥A k ∥. Set

$$\displaystyle{\mu = \frac{\mu _{1}} {\mathcal{L} + Q}}$$

and consider d so small that inequality (2.52) holds for | v | ≤ δ with \(\delta = (\mathcal{L} + Q)d\).

The definition of the vectors t k implies that

$$\displaystyle{dt_{k+1} = y_{k+1} - p_{k+1} = f(\,y_{k}) - f(\,p_{k}) = f(\,p_{k} + dt_{k}) - f(\,p_{k}).}$$

Since \(\vert dt_{k}\vert \leq (\mathcal{L} + Q)d\) by (2.59), it follows from Condition S and from the above choice of d that

$$\displaystyle{\left \vert dt_{k+1} - dA_{k}t_{k}\right \vert = \left \vert \,f(\,p_{k} + dt_{k}) - f(\,p_{k}) - dA_{k}t_{k}\right \vert \leq }$$
$$\displaystyle{\leq \mu \vert dt_{k}\vert \leq \mu (\mathcal{L} + Q)d =\mu _{1}d.}$$

Hence,

$$\displaystyle{ t_{k+1} = A_{k}t_{k} +\theta _{k},\quad \mbox{ where}\quad \vert \theta _{k}\vert <\mu _{1}. }$$
(2.62)

Consider the vectors

$$\displaystyle{c_{k} = t_{k} - b_{k}.}$$

Note that c N = 0 by (2.60) and

$$\displaystyle{c_{k+1} = A_{k}c_{k} +\theta _{k},\quad \mbox{ where}\quad \vert \theta _{k}\vert <\mu _{1}}$$

by (2.62).

Thus,

$$\displaystyle{\vert c_{-N+1}\vert \leq \vert \theta _{-N}\vert <\mu _{1},}$$
$$\displaystyle{\vert c_{-N+2}\vert \leq \vert A_{-N+1}c_{-N+1} +\theta _{-N+1}\vert \leq (K + 1)\mu _{1},}$$

and so on, which implies the estimate

$$\displaystyle{\vert c_{k}\vert \leq \left ((K + 1)^{2N} + (K + 1)^{2N-1} +\ldots +1\right )\mu _{ 1} < 1,\quad - N \leq k \leq N.}$$

Hence,

$$\displaystyle{ \vert t_{k} - b_{k}\vert \leq 1,\quad - N \leq k \leq N. }$$
(2.63)

Finally, we consider the sequence

$$\displaystyle{z_{k} = \left \{\begin{array}{ll} 0, &k < -N; \\ \varDelta _{k} - b_{k},& - N \leq k \leq N + 1; \\ 0, &k > N + 1.\\ \end{array} \right.}$$

Relations (2.55) and (2.60) imply relations (2.54); estimates (2.58) and (2.63) imply estimate (2.53). □

Lemma 2.3.2

The sequence \(\mathcal{A} =\{ A_{k}\}\) has the Perron property.

Proof

Take an arbitrary sequence

$$\displaystyle{w_{k} \in \mathbb{R}^{n},\quad k \in \mathbb{Z},}$$

with | w k | < 1 and prove that an analog of Eq. (2.54) has a solution

$$\displaystyle{z_{k} \in \mathbb{R}^{n},\quad k \in \mathbb{Z},}$$

with

$$\displaystyle{\vert z_{k}\vert \leq \mathcal{L} + 1,\quad k \in \mathbb{Z}.}$$

Fix a natural N and consider the sequence

$$\displaystyle{w_{k}^{(N)} = \left \{\begin{array}{ll} w_{k},&\ - N \leq k \leq N; \\ 0, &\vert k\vert \geq N + 1.\\ \end{array} \right.}$$

By Lemma 2.3.1, there exists a sequence \(\left \{z_{k}^{(N)},\;k \in \mathbb{Z}\right \}\) such that

$$\displaystyle{ z_{k+1}^{(N)} = A_{ k}z_{k}^{(N)} + w_{ k}^{(N)},\quad - N \leq k \leq N, }$$
(2.64)

and

$$\displaystyle{ \left \vert z_{k}^{(N)}\right \vert \leq \mathcal{L} + 1,\quad k \in \mathbb{Z}. }$$
(2.65)

Passing to a subsequence of \(\left \{z_{k}^{(N)}\right \}\), we can find a sequence {v k } such that

$$\displaystyle{v_{k} =\lim _{N\rightarrow \infty }z_{k}^{(N)},\quad k \in \mathbb{Z}.}$$

(Note that do not assume uniform convergence.) Passing to the limit in (2.64) and (2.65) as N, we see that

$$\displaystyle{v_{k+1} = A_{k}y_{k} + w_{k},\quad k \in \mathbb{Z},}$$

and

$$\displaystyle{\vert v_{k}\vert \leq \mathcal{L} + 1,\quad k \in \mathbb{Z}.}$$

Thus, we have shown that the sequence \(\mathcal{A}\) has the Perron property. □

Now let us explain how to prove the required statement in the case of a smooth closed manifold M.

Lemma 2.3.3

If a diffeomorphism of class C 1 of a smooth closed n-dimensional manifold M has the Lipschitz shadowing property, \(\left \{\,p_{k} = f^{k}(\,p)\right \}\) is an arbitrary trajectory of f, and A k = Df( p k ), then the sequence \(\mathcal{A} =\{ A_{k}\}\) has the Perron property.

Proof

Let exp be the standard exponential mapping on the tangent bundle of M generated by the fixed Riemannian metric dist. Let

$$\displaystyle{\exp _{x}: T_{x}M \rightarrow M}$$

be the corresponding exponential mapping at a point xM.

Denote (just for this proof) by B(r, x) the ball in M of radius r centered at a point x; let B T (r, x) be the ball in T x M of radius r centered at the origin.

It is well known that there exists an r > 0 such that for any xM, exp x is a diffeomorphism of B T (r, x) onto its image and exp x −1 is a diffeomorphism of B(r, x) onto its image; in addition, Dexp x (0) = Id.

Thus, we may assume that r is chosen so that the following inequalities hold for any xM:

$$\displaystyle{ \mbox{ dist}(\exp _{x}(v),\exp _{x}(w)) \leq 2\vert v - w\vert,\quad v,w \in B_{T}(r,x), }$$
(2.66)

and

$$\displaystyle{ \left \vert \exp _{x}^{-1}(\,y) -\exp _{ x}^{-1}(z)\right \vert \leq 2\mbox{ dist}(\,y,z),\quad y,z \in B(r,x). }$$
(2.67)

These inequalities mean that distances are distorted not more than twice when we pass from the manifold to its tangent space or from the tangent space to the manifold (if we work in a small neighborhood of a point of the manifold or in a small neighborhood of the origin of the tangent space).

In our reasoning below, we always assume that d is so small that the corresponding points belong to such small neighborhoods.

Now we fix a trajectory \(\left \{\,p_{k} = f^{k}(\,p)\right \}\) of our diffeomorphism f and introduce the mappings

$$\displaystyle{\,F_{k} =\exp _{ p_{k+1}}^{-1} \circ f \circ \exp _{ p_{k}}:\; T_{p_{k}}M \rightarrow T_{p_{k+1}}M.}$$

Clearly,

$$\displaystyle{DF_{k}(0) = A_{k}.}$$

The analog of Condition S is as follows: For any μ > 0 we can find a δ > 0 (independent of k) such that if | v | < δ, then

$$\displaystyle{ \vert \,F_{k}(v) - A_{k}v\vert \leq \mu \vert v\vert,\quad k \in \mathbb{Z}. }$$
(2.68)

Of course, this condition is satisfied automatically since f is of class C 1 and the manifold M is compact.

To prove that the sequence \(\mathcal{A}\) has the Perron property, let us consider the difference equations

$$\displaystyle{ v_{k+1} = A_{k}y_{k} + w_{k},\quad k \in \mathbb{Z}, }$$
(2.69)

where \(v_{k} \in T_{p_{k}}M\) and \(w_{k} \in T_{p_{k+1}}M\).

We assume that \(\vert w_{k}\vert < 1,\;k \in \mathbb{Z}\). Let us “translate” the reasoning of Lemma 2.3.1 to the “manifold language.”

We fix a natural N and consider the sequence

$$\displaystyle{\varDelta _{k} \in T_{p_{k}}M,\quad - N \leq k \leq N + 1,}$$

defined by relations (2.55). Let Q satisfy (2.56).

We fix a small d and define the sequence \(\xi =\{ x_{k} \in M:\; k \in \mathbb{Z}\}\) by

$$\displaystyle{x_{k} = \left \{\begin{array}{ll} f^{k+N}(\,p_{-N}), &k < -N; \\ \exp _{p_{k}}(d\varDelta _{k}), & - N \leq k \leq N + 1; \\ f^{k-N-1}(\exp _{p_{N+1}}(d\varDelta _{N+1})),&k > N + 1.\\ \end{array} \right.}$$

This definition and inequalities (2.66) imply that if d is small enough, then

$$\displaystyle{\mbox{ dist}\left (x_{k+1},\exp _{p_{k+1}}(dA_{k}\varDelta _{k})\right ) < 2d.}$$

Since

$$\displaystyle{\,f(x_{k}) =\exp _{p_{k+1}}(\,F_{k}(d\varDelta _{k})),}$$

condition (2.68) with μ < 1 implies that

$$\displaystyle{\mbox{ dist}\left (\exp _{p_{k+1}}(dA_{k}\varDelta _{k}),f(x_{k})\right ) < 2d,}$$

and we see that

$$\displaystyle{\mbox{ dist}\left (\,f(x_{k}),x_{k+1}\right ) < 4d.}$$

Thus, there exists an exact trajectory \(\eta =\{ y_{k}:\; k \in \mathbb{Z}\}\) of f such that

$$\displaystyle{ \mbox{ dist}(x_{k},y_{k}) \leq 4\mathcal{L}d,\quad k \in \mathbb{Z}. }$$
(2.70)

Now we consider the finite sequence

$$\displaystyle{t_{k} = \frac{1} {d}\exp _{p_{k}}^{-1}(\,y_{ k}),\quad - N \leq k \leq N.}$$

Inequalities (2.70) and (2.67) imply that

$$\displaystyle{ \vert \varDelta _{k} - t_{k}\vert \leq 8\mathcal{L},\quad k \in \mathbb{Z}. }$$
(2.71)

Note that

$$\displaystyle{\mbox{ dist}(\,y_{k},p_{k}) \leq \mbox{ dist}(\,y_{k},x_{k}) + \mbox{ dist}(x_{k},p_{k}) \leq (4\mathcal{L} + 2Q)d,\quad k \in \mathbb{Z}.}$$

Hence,

$$\displaystyle{\vert t_{k}\vert \leq 8\mathcal{L} + 4Q,\quad k \in \mathbb{Z}.}$$

Now we define a finite sequence

$$\displaystyle{b_{k} \in T_{p_{k}}M,\quad - N \leq k \leq N + 1,}$$

by relations (2.60) and repeat the reasoning of Lemma 2.3.1 with

$$\displaystyle{\mu = \frac{\mu _{1}} {8\mathcal{L} + 4Q},}$$

where μ 1 is the same as above (see relation (2.61)).

The rest of the proof is literally the same (with natural replacement of \(\mathbb{R}^{n}\) by the corresponding tangent spaces), and we get the relation

$$\displaystyle{\vert t_{k} - b_{k}\vert < 1}$$

similar to (2.63).

Finally, we get the estimate

$$\displaystyle{\vert z_{k}\vert \leq 8\mathcal{L} + 1,}$$

which completes the proof of the analog of Lemma 2.3.1.

The rest of the proof of the implication “Lipschitz shadowing property implies the Perron property of the sequence \(\mathcal{A}\)” almost literally repeats the proof of Lemma 2.3.2. □

Historical Remarks

Theorem 2.3.1 was published by the first author and S. B. Tikhomirov in the paper [68]. Let us mention that the paper [67] contained the first proof of the fact that structural stability follows from certain shadowing property based on a combination of the Maizel’, Pliss, and Mañé theorems.

2.4 Lipschitz Periodic Shadowing for Diffeomorphisms

The main result of this section is as follows.

Theorem 2.4.1

A diffeomorphism f of class C 1 of a smooth closed n-dimensional manifold M has the Lipschitz periodic shadowing property if and only if f is Ω-stable.

First we prove the “if” statement of Theorem 2.4.1.

Theorem 2.4.2

If a diffeomorphism f is Ω-stable, then f has the Lipschitz periodic shadowing property.

Let us give one more definition.

Definition 2.4.1

We say that a diffeomorphism f has the Lipschitz shadowing property on a set U if there exist positive constants \(\mathcal{L},d_{0}\) such that if \(\xi =\{ x_{i}:\; i \in \mathbb{Z}\} \subset U\) is a d-pseudotrajectory with dd 0, then there exists a point pU such that inequalities (1.5) hold.

Remark 2.4.1

It follows from Theorems 1.4.2 and 2.2.4 that we can find a neighborhood U of a hyperbolic set Λ of a diffeomorphism f having the above-formulated property and such that f is expansive on U.

We start by proving several auxiliary results.

Lemma 2.4.1

Let f be a homeomorpism of a compact metric space (M, dist). For any neighborhood U of the nonwandering set Ω( f) there exist positive numbers T, δ 1 such that if \(\xi =\{ x_{i}:\; i \in \mathbb{Z}\}\) is a d-pseudotrajectory of f with dδ 1 and

$$\displaystyle{x_{k},x_{k+1},\ldots,x_{k+l}\notin U}$$

for some \(k \in \mathbb{Z}\) and l > 0, then lT.

Proof

Take a neighborhood U of the nonwandering set Ω( f) and let T be the Birkhoff constant for the homeomorphism f given for this neighborhood by Theorem 2.2.5. Assume that there does not exist a number δ 1 with the desired property; then there exists a sequence d j → 0 as j and a sequence of d j -pseudotrajectories \(\{x_{k}^{(\,j)}:\; k \in \mathbb{Z}\}\) of f such that

$$\displaystyle{\left \{x_{k}^{(\,j)}:\; 0 \leq k \leq T - 1\right \} \cap U =\emptyset }$$

for all j.

The set M′ = M∖U is compact. Passing to a subsequence, if necessary, we may assume that x 0 ( j)x 0 as j. In this case,

$$\displaystyle{x_{k}^{(\,j)} \rightarrow f^{k}(x_{ 0}) \in M',\quad 0 \leq k \leq T - 1,}$$

and we get a contradiction with the choice of T. □

Now let us recall some basic properties of Ω-stable diffeomorphisms. It was noted in Sect. 1.3 that a diffeomorphism f is Ω-stable if and only if f satisfies Axiom A and the no cycle condition (Theorem 1.3.3).

Let Ω 1, , Ω m be the basic sets in decomposition (1.15) of the nonwandering set of an Ω-stable diffeomorphism f.

Below we need one folklore technical statement. Recall that we write Ω i Ω j if there is a point xΩ( f) such that

$$\displaystyle{\,f^{-k}(x) \rightarrow \varOmega _{ i}\mbox{ and }f^{k}(x) \rightarrow \varOmega _{ j},\quad k \rightarrow \infty.}$$

Theorem 2.4.3

Assume that a diffeomorphism f is Ω-stable. For any family of neighborhoods U i of the basic sets Ω i one can find neighborhoods V i U i such that if a point x belongs to some V i and there exist indices 0 < lm such that

$$\displaystyle{\,f^{l}(x)\notin U_{ i}\mathit{\mbox{ and }}f^{m}(x) \in V _{ j},}$$

then there exist basic sets \(\varOmega _{i_{1}},\ldots,\varOmega _{i_{t}}\) such that

$$\displaystyle{ \varOmega _{i} \rightarrow \varOmega _{i_{1}} \rightarrow \ldots \rightarrow \varOmega _{i_{t}} \rightarrow \varOmega _{j}. }$$
(2.72)

Proof

Reducing the given neighborhoods U i , we may assume that the compact sets U i = f(Cl(U i )) ∪Cl(U i ) are disjoint.

Assume that our statement does not hold. In this case, there exist sequences of points x k , k ≥ 0, and indices l(k) ≤ m(k) such that

$$\displaystyle{x_{k} \rightarrow \varOmega _{i},\quad f^{l(k)}(x_{ k})\notin U_{i},\quad f^{m(k)}(x_{ k}) \rightarrow \varOmega _{i},\quad k \rightarrow \infty.}$$

Clearly, we may assume that

$$\displaystyle{x_{k},f(x_{k}),\ldots,f^{l(k)-1}(x_{ k}) \in U_{i}}$$

while

$$\displaystyle{\,y_{k}:= f^{l(k)}(x_{ k})\notin U_{i}.}$$

Then y k U i , and, passing to a subsequence, if necessary, we may assume that y k yU i as k.

Since Ω i is a compact f-invariant set, l(k) → as k. Thus, for any t < 0, f t( y k ) ∈ U i for large k, and it follows that f t( y) ∈ Cl(U i ) for any t < 0. We note that the set Cl(U i ) intersects a single basic set, Ω i , and refer to (1.16) to conclude that

$$\displaystyle{ y \in W^{u}(\varOmega _{ i}). }$$
(2.73)

By the same relation (1.16), there exists a basic set \(\varOmega _{i_{1}}\) such that

$$\displaystyle{ y \in W^{s}(\varOmega _{ i_{1}}). }$$
(2.74)

By our choice of U i , the sets Cl( f(U i ))∖U i do not contain nonwandering points. Thus, if i 1 = i, inclusions (2.73) and (2.74) mean the existence of a 1-cycle, and we get the desired contradiction.

Hence, i 1i and \(\varOmega _{i} \rightarrow \varOmega _{i_{1}}\). Consider the compact set

$$\displaystyle{\,Y = \left \{\,f^{k}(\,y):\, k \geq 0\right \} \cup \varOmega _{ i_{1}}.}$$

Clearly, the set Y has a neighborhood Z such that \(U_{i_{1}} \subset Z\) and Z does not intersect a small neighborhood of Ω i .

Since \(y_{k} = f^{l_{k}}(x_{ k}) \rightarrow y\), there exist indices l 1(k) such that

$$\displaystyle{\,f^{t}(\,y_{ k}) = f^{l(k)+t}(x_{ k}) \in Z,\quad 0 \leq t \leq l_{1}(k),}$$

for large k, and

$$\displaystyle{x_{1,k} = f^{l_{1}(k)}(\,y_{ k}) = f^{l(k)+l_{1}(k)}(x_{ k}) \rightarrow \varOmega _{i_{1}},\quad k \rightarrow \infty.}$$

At the same time, the positive trajectories of the points y k (and hence, of the points x 1,k ) must leave Z (and hence, \(U_{i_{1}}\)) since the sequence

$$\displaystyle{\,f^{m(k)-l(k)}(\,y_{ k}) = f^{m(k)}(x_{ k})}$$

tends to Ω i .

Thus, we can repeat the above reasoning with the points x 1,k and the basic set \(\varOmega _{i_{1}}\) instead of x k and Ω i .

Such a process will produce basic sets \(\varOmega _{i_{1}},\varOmega _{i_{2}},\ldots\) such that

$$\displaystyle{\varOmega _{i} \rightarrow \varOmega _{i_{1}} \rightarrow \varOmega _{i_{2}} \rightarrow \ldots.}$$

Since f has no cycles, this process is finite, and, as a result, we conclude that there exist basic sets \(\varOmega _{i_{1}},\ldots,\varOmega _{i_{t}}\) such that relations (2.72) hold. □

Now we apply the above theorem to prove a statement concerning periodic pseudotrajectories of Ω-stable diffeomorphisms.

Lemma 2.4.2

Assume that a diffeomorphism f is Ω-stable. For any family of disjoint neighborhoods W i of the basic sets Ω i there exists a number δ 2 > 0 such that any periodic d-pseudotrjectory ξ of f with dδ 2 belongs to a single neighborhood W i .

Proof

Fix arbitrary disjoint neighborhoods W i of the basic sets Ω i and find a number ɛ > 0 and neighborhoods U i of Ω i such that

$$\displaystyle{N(\varepsilon,U_{i}) \subset W_{i},\quad i = 1,\ldots,m.}$$

Apply Theorem 2.4.3 to find for U i the corresponding neighborhoods V i of Ω i . Reducing ɛ, if necessary, we can find neighborhoods V i of Ω i such that

$$\displaystyle{N(\varepsilon,V _{i}^{{\prime}}) \subset V _{ i},\quad i = 1,\ldots,m.}$$

By Lemma 2.4.1, there exist positive numbers T, δ 1 such that if ξ = {x k } is a d-pseudotrajectory of f with dδ 1 and

$$\displaystyle{x_{k},x_{k+1},\ldots,x_{k+l}\notin V:=\bigcup _{ i=1}^{m}V _{ i}^{{\prime}}}$$

for some \(k \in \mathbb{Z}\) and l > 0, then lT.

Find a number δ 2 ∈ (0, δ 1) such that if ξ = {x k } is a d-pseudotrajectory of f with dδ 2, then

$$\displaystyle{\mbox{ dist}(\,f^{l}(x_{ k}),x_{k+l}) <\varepsilon,\quad 0 \leq l \leq T + 1,}$$

for any \(k \in \mathbb{Z}\).

Now let ξ = {x k } be a periodic d-pseudotrajectory of f of period μ with dδ 2.

Let us call a V -block of ξ a finite segment

$$\displaystyle{\xi _{k,m} =\{ x_{k},x_{k+1},\ldots,x_{k+m}\},\quad k \in \mathbb{Z},\;m > 0,}$$

such that x k , x k+m V while x k+l V for 0 < l < m. Note that in this case, mT + 1.

Let us note simple properties of V -blocks.

It follows from the choice of δ 2 that if ξ k, m is a V -block for which there exist indices i, j ∈ {1, , m} such that x k V i and x k+m V j , then dist( f m(x k ), x k+m ) < ɛ; hence, f m(x k ) ∈ V j .

At the same time, if for such a V -block there exists an index l ∈ (0, m) such that x k+l W i , then dist( f l(x k ), x k+l ) < ɛ; hence, f l(x k ) ∉ U i .

It follows from Theorem 2.4.3 that in this case, there exists a relation of the form (2.72); the absence of cycles implies that ji.

Since δ 2 < δ 1, there exists a neighborhood V i such that ξ intersects V i .

Changing indices of ξ, we may assume that x 0V i .

If either x k W i for k ≥ 0 or any V -block ξ k, m with k ≥ 0 belongs to W i , then the statement of our lemma follows from the periodicity of ξ.

It was noted above that if ξ k, m be a V -block with x k V j for k ≥ 0 for which there exists an index l ∈ (0, m) such that \(x_{k_{l}}\notin W_{j}\), then there exists an index j′ ≠ j for which we have a relation

$$\displaystyle{\varOmega _{j} \rightarrow \ldots \rightarrow \varOmega _{j'}}$$

of the form (2.72).

Thus, if we assume that there exists a V -block ξ k, m with k ≥ 0 such that ξ k, m ∖W i ≠ ∅, then we get an index j 1i such that we have a relation

$$\displaystyle{\varOmega _{i} \rightarrow \ldots \rightarrow \varOmega _{j_{1}}}$$

of the form (2.72).

Going to “the right” of this V -block ξ k, m and continuing this process, we construct a sequence of pairs of indices (i, j 1), ( j 1, j 2),  such that

$$\displaystyle{\varOmega _{i} \rightarrow \ldots \rightarrow \varOmega _{j_{1}},\quad \varOmega _{j_{1}} \rightarrow \ldots \rightarrow \varOmega _{j_{2}},\quad \ldots.}$$

In this case, it follows from the absence of cycles that all the indices i, j 1, j 2,  are different.

But the μ-periodicity of ξ implies that if ξ k, m is a V -block and n is a natural number, then ξ k+, m is an identical V -block, and the existence of the above sequence with different i, j 1, j 2,  is impossible.

Now we prove Theorem 2.4.2.

By Remark 2.4.1, there exist disjoint neighborhoods U 1, , U m of the basic sets Ω 1, , Ω m such that

  1. (i)

    f has the Lipschitz shadowing property on any of U j with the same constants \(\mathcal{L},d_{0}^{{\ast}}\);

  2. (ii)

    f is expansive on any of U j with the same expansivity constant a.

Find neighborhoods W j of Ω j (and reduce d 0 , if necessary) so that the \(\mathcal{L}d_{0}^{{\ast}}\)-neighborhoods of W j belong to U j . Apply Lemma 2.4.2 to find the corresponding constant δ 2.

We claim that f has the Lipschitz periodic shadowing property with constants \(\mathcal{L},d_{0}\), where

$$\displaystyle{d_{0} =\min \left (d_{0}^{{\ast}},\delta _{ 2}, \frac{a} {2\mathcal{L}}\right ).}$$

Take a μ-periodic d-pseudotrajectory ξ = {x k } of f with dd 0. Lemma 2.4.2 implies that there exists a neighborhood W i such that ξW i U i .

Thus, there exists a point p such that inequalities (1.5) hold. Let us show that p is a periodic point of f. By the choice of U i and W i , f k( p) ∈ U i for all \(k \in \mathbb{Z}\). Let q = f μ( p). Inequalities (1.5) and the periodicity of ξ imply that

$$\displaystyle{\mbox{ dist}\left (\,f^{k}(q),x_{ k}\right ) = \mbox{ dist}\left (\,f^{k+\mu }(\,p),x_{ k}\right ) = \mbox{ dist}\left (\,f^{k+\mu }(\,p),x_{ k+\mu }\right ) \leq \mathcal{L}d,\quad k \in \mathbb{Z}.}$$

Thus,

$$\displaystyle{\mbox{ dist}\left (\,f^{k}(q),f^{k}(\,p)\right ) \leq 2\mathcal{L}d \leq a,\quad k \in \mathbb{Z},}$$

which implies that f μ( p) = q = p. This completes the proof. □

Now we prove the “only if” statement of Theorem 2.4.1.

Theorem 2.4.4

If a diffeomorphism f has the Lipschitz periodic shadowing property, then f is Ω-stable.

Thus, let us assume that f has the Lipschitz periodic shadowing property (with constants \(\mathcal{L} \geq 1,d_{0} > 0\)). Clearly, in this case f −1 has the Lipschitz periodic shadowing property as well (and we assume that the constants \(\mathcal{L},d_{0}\) are the same for f and f −1).

To clarify the presentation, in the construction of pseudotrajectories in the following Lemmas 2.4.3 and 2.4.4, we assume that f is a diffeomorphism of \(\mathbb{R}^{n}\) (and leave to the reader consideration of the case of a manifold).

We also assume that there exists a number N > 0 such that ∥Df(x)∥ ≤ N for all considered points x (an analog of this assumption is satisfied in the case of a closed manifold).

Recall that we denote by Per( f) the set of periodic points of f.

Lemma 2.4.3

Every point pPer( f) is hyperbolic.

Proof

To get a contradiction, let us assume that f has a nonhyperbolic periodic point p (to simplify notation, we assume that p is a fixed point; literally the same reasoning can be applied to a periodic point of period m > 1). In addition, we assume that p = 0.

In this case, we can represent

$$\displaystyle{\,f(v) = Av + F(v),}$$

where A = Df(0) and F(v) = o(v) as v → 0.

By our assumption, A is a nonhyperbolic matrix. The following two cases are possible:

  • Case 1: A has a real eigenvalue λ with | λ | = 1;

  • Case 2: A has a complex eigenvalue λ with | λ | = 1.

We treat in detail only Case 1 and give a comment concerning Case 2. To simplify presentation, we assume that 1 is an eigenvalue of A; the case of eigenvalue − 1 is treated similarly.

We can introduce coordinate v such that, with respect to this coordinate, the matrix A has block-diagonal form,

$$\displaystyle{ A = \mbox{ diag}(B,P), }$$
(2.75)

where B is a Jordan block of size l × l:

$$\displaystyle{B = \left (\begin{array}{ccccc} 1&1&0&\ldots &0\\ 0 &1 &1 &\ldots &0\\ \vdots & \vdots & \vdots &\ddots & \vdots \\ 0&0&0&\ldots &1 \end{array} \right ).}$$

Of course, introducing new coordinates, we have to change the constants \(\mathcal{L}\) and d 0; we denote the new constants by the same symbols. In addition, we assume that \(\mathcal{L}\) is integer.

We start considering the case l = 2; in this case,

$$\displaystyle{B = \left (\begin{array}{cc} 1&1\\ 0 &1 \end{array} \right ).}$$

Let

$$\displaystyle{e_{1} = (1,0,0,\ldots,0)\mbox{ and }e_{2} = (0,1,0,\ldots,0)}$$

be the first two vectors of the standard orthonormal basis.

Let \(K = 7\mathcal{L}\).

Take a small d > 0 and construct a finite sequence y 0, , y Q of points (where Q is determined later) as follows: y 0 = 0 and

$$\displaystyle{ y_{k+1} = Ay_{k} + de_{2},\quad k = 0,\ldots,K - 1. }$$
(2.76)

Then

$$\displaystyle{\,y_{K} = (Z_{1}(K)d,Kd,0,\ldots,0),}$$

where the natural number Z 1(K) is determined by K (we do not write Z 1(K) explicitly). Now we set

$$\displaystyle{\,y_{k+1} = Ay_{k} - de_{2},\quad k = K,\ldots,2K - 1.}$$

Then

$$\displaystyle{\,y_{2K} = (Z_{2}(K)d,0,0,\ldots,0),}$$

where the natural number Z 2(K) is determined by K as well. Take Q = 2K + Z 2(K); if we set

$$\displaystyle{\,y_{k+1} = Ay_{k} - de_{1},\quad k = 2K,\ldots,Q - 1,}$$

then y Q = 0. Let us note that both numbers Q and

$$\displaystyle{\,Y:= \frac{\max _{0\leq k\leq Q-1}\vert \,y_{k}\vert } {d} }$$

are determined by K (and hence, by \(\mathcal{L}\)).

Now we construct a Q-periodic sequence \(x_{k},k \in \mathbb{Z},\) that coincides with the above sequence for k = 0, , Q.

We claim that if d is small enough, then ξ = {x k } is a 2d-pseudotrajectory of f (and this pseudotrajectory is Q-periodic by construction).

Indeed, we know that | x k | ≤ Yd for \(k \in \mathbb{Z}\). Since F(v) = o( | v | ) as | v | → 0,

$$\displaystyle{ \vert \,F(x_{k})\vert < d,\quad k \in \mathbb{Z}, }$$
(2.77)

if d is small enough.

The definition of x k implies that

$$\displaystyle{ \vert x_{k+1} - Ax_{k}\vert = d,\quad k \in \mathbb{Z}. }$$
(2.78)

It follows from (2.77) and (2.78) that

$$\displaystyle{\vert x_{k+1} - f(x_{k})\vert \leq \vert x_{k+1} - Ax_{k}\vert + \vert \,F(x_{k})\vert < 2d,}$$

which implies that ξ = {x k } is a 2d-pseudotrajectory of f if d is small enough.

Now we estimate the distances between points of trajectories of the diffeomorphism f and its linearization at zero.

Let us take a vector p 0 and assume that the sequence p k = f k( p 0) belongs to the ball \(\vert v\vert \leq (Y + 2\mathcal{L})d\) for 0 ≤ kK. Let r k = A k p 0 (we impose no conditions on r k since below we estimate F at points q k only).

Take a small number μ ∈ (0, 1) (to be chosen later) and assume that d is small enough, so that the inequality

$$\displaystyle{\vert \,F(v)\vert \leq \mu \vert v\vert }$$

holds for \(\vert v\vert \leq (Y + 2\mathcal{L})d\).

By our assumption, ∥A∥ = ∥Df(0)∥ ≤ N. Then

$$\displaystyle{\vert \,p_{1}\vert \leq \vert Ap_{0}\vert + \vert \,F(\,p_{0})\vert \leq (N + 1)\vert \,p_{0}\vert,\ldots,}$$
$$\displaystyle{\vert \,p_{k}\vert \leq \vert Ap_{k-1}\vert + \vert \,F(\,p_{k-1})\vert \leq (N + 1)^{k}\vert \,p_{ 0}\vert }$$

for 1 ≤ kK, and

$$\displaystyle{\vert \,p_{1} - r_{1}\vert = \vert Ap_{0} + F(\,p_{0}) - Ap_{0}\vert \leq \mu \vert \,p_{0}\vert,}$$
$$\displaystyle{\vert \,p_{2} - r_{2}\vert = \vert Ap_{1} + F(\,p_{1}) - Ar_{1}\vert \leq N\vert \,p_{1} - r_{1}\vert +\mu \vert \,p_{1}\vert \leq \mu (2N + 1)\vert \,p_{0}\vert,}$$
$$\displaystyle{\vert \,p_{3} - r_{3}\vert \leq N\vert \,p_{2} - r_{2}\vert +\mu \vert \,p_{2}\vert \leq \mu (N(2N + 1) + (N + 1)^{2})\vert \,p_{ 0}\vert,}$$

and so on.

Thus, there exists a number ν = ν(K, N) such that

$$\displaystyle{\vert \,p_{k} - r_{k}\vert \leq \mu \nu \vert \,p_{0}\vert,\quad 0 \leq k \leq K.}$$

We take μ = 1∕ν, note that μ = μ(K, N), and get the inequalities

$$\displaystyle{ \vert \,p_{k} - r_{k}\vert \leq \vert \,p_{0}\vert,\quad 0 \leq k \leq K, }$$
(2.79)

for d small enough.

Since f has the Lipschitz periodic shadowing property, for d small enough, the Q-periodic 2d-pseudotrajectory ξ is \(2\mathcal{L}d\)-shadowed by a periodic trajectory. Let p 0 be a point of this trajectory such that

$$\displaystyle{ \vert \,p_{k} - x_{k}\vert \leq \mathcal{L}d,\quad k \in \mathbb{Z}, }$$
(2.80)

where p k = f k( p 0).

The inequalities | x k | ≤ Yd and (2.80) imply that

$$\displaystyle{ \vert \,p_{k}\vert \leq \vert x_{k}\vert + \vert \,p_{k} - x_{k}\vert \leq (\,Y + 2\mathcal{L})d,\quad k \in \mathbb{Z}. }$$
(2.81)

Note that \(\vert \,p_{0}\vert \leq 2\mathcal{L}d\).

Set r k = A k p 0; we deduce from estimate (2.79) that if d is small enough, then

$$\displaystyle{ \vert \,p_{K} - r_{K}\vert \leq \vert \,p_{0}\vert \leq 2\mathcal{L}d. }$$
(2.82)

Denote by v (2) the second coordinate of a vector v.

It follows from the structure of the matrix A that

$$\displaystyle{ \left \vert r_{K}^{(2)}\right \vert = \left \vert \,p_{ 0}^{(2)}\right \vert \leq 2\mathcal{L}d. }$$
(2.83)

The relations

$$\displaystyle{\left \vert \,y_{K}^{(2)}\right \vert = Kd\mbox{ and }\left \vert \,p_{ K} - y_{K}\right \vert \leq 2\mathcal{L}d}$$

imply that

$$\displaystyle{ \left \vert \,p_{K}^{(2)}\right \vert \geq Kd - 2\mathcal{L}d = 5\mathcal{L}d }$$
(2.84)

(recall that \(K = 7\mathcal{L}\)).

Estimates (2.82)–(2.84) are contradictory. Our lemma is proved in Case 1 for l = 2.

If l = 1, then the proof is simpler; the first coordinate of A k v equals the first coordinate of v, and we construct the periodic pseudotrajectory perturbing the first coordinate only.

If l > 2, the reasoning is parallel to that above; we first perturb the lth coordinate to make it Kd, and then produce a periodic sequence consequently making zero the lth coordinate, the (l − 1)st coordinate, and so on.

If λ is a complex eigenvalue, λ = a + bi, we take a real 2 × 2 matrix

$$\displaystyle{R = \left (\begin{array}{cc} a& - b\\ b & a\\ \end{array} \right )}$$

and assume that in representation (2.75), B is a real 2l × 2l Jordan block:

$$\displaystyle{B = \left (\begin{array}{ccccc} R&E_{2} & 0 &\ldots & 0 \\ 0 & R &E_{2} & \ldots & 0\\ \vdots & \vdots & \vdots &\ddots & \vdots \\ 0 & 0 & 0 &\ldots &R \end{array} \right ),}$$

where E 2 is the 2 × 2 identity matrix.

After that, almost the same reasoning works; we note that | Rv | = | v | for any 2-dimensional vector v and construct periodic pseudotrajectories replacing, for example, formulas (2.76) by the formulas

$$\displaystyle{\,y_{k+1} = Ay_{k} + dw_{k},\quad k = 0,\ldots,K - 1,}$$

where jth coordinates of the vector w k are zero for j = 1, , 2l − 2, 2l + 1, , n, while the 2-dimensional vector corresponding to (2l − 1)st and 2lth coordinates has the form R k w with | w | = 1, and so on. We leave details to the reader. The lemma is proved. □

Lemma 2.4.4

There exist constants C > 0 and λ ∈ (0, 1) depending only on N and \(\mathcal{L}\) and such that, for any point pPer( f), there exist complementary subspaces S( p) and U( p) of \(\mathbb{R}^{n}\) that are Df-invariant, i.e.,

  • (H1) Df( p)S( p) = S( f( p)) and Df( p)U( p) = U( f( p)),

    and the inequalities

  • (H2.1) \(\left \vert Df^{j}(\,p)v\right \vert \leq C\lambda ^{j}\vert v\vert,\quad v \in S(\,p),j \geq 0\) ,

    and

  • (H2.2) \(\left \vert Df^{-j}(\,p)v\right \vert \leq C\lambda ^{j}\vert v\vert,\quad v \in U(\,p),j \geq 0\) ,

    hold.

Remark 2.4.2

This lemma means that the set Per( f) has all the standard properties of a hyperbolic set, with the exception of compactness.

Proof

Take a periodic point p ∈ Per( f); let m be the minimal period of p.

Denote p i = f i( p), A i = Df( p i ), and B = Df m( p). It follows from Lemma 2.4.3 that the matrix B is hyperbolic. Denote by S( p) and U( p) the invariant subspaces of B corresponding to parts of its spectrum inside and outside the unit disk, respectively. Clearly, S( p) and U( p) are invariant with respect to Df, they are complementary subspaces of \(\mathbb{R}^{n}\), and the following relations hold:

$$\displaystyle{ \lim _{n\rightarrow +\infty }B^{n}v_{ s} =\lim _{n\rightarrow +\infty }B^{-n}v_{ u} = 0,\quad v_{s} \in S(\,p),v_{u} \in U(\,p). }$$
(2.85)

We prove that inequalities (H2.2) hold with \(C = 4\mathcal{L}\) and \(\lambda = 1 + 1/(2\mathcal{L})\) (inequalities (H2.1) are established by similar reasoning applied to f −1 instead of f).

Consider an arbitrary nonzero vector v u U( p) and an integer j ≥ 0. Define sequences of vectors v i , e i and numbers λ i > 0 for i ≥ 0 as follows:

$$\displaystyle{v_{0} = v_{u},\quad v_{i+1} = A_{i}v_{i},\quad e_{i} = \frac{v_{i}} {\vert v_{i}\vert },\quad \lambda _{i} = \frac{\vert v_{i+1}\vert } {\vert v_{i}\vert } = \vert A_{i}e_{i}\vert.}$$

Let

$$\displaystyle{\tau = \frac{\lambda _{m-1} \cdot \ldots \cdot \lambda _{1} + \lambda _{m-1} \cdot \ldots \cdot \lambda _{2} +\ldots +\lambda _{m-1} + 1} {\lambda _{m-1} \cdot \ldots \cdot \lambda _{0}}.}$$

Consider the sequence \(\{a_{i} \in \mathbb{R}:\; i \geq 0\}\) defined by the following formulas:

$$\displaystyle{ a_{0} =\tau,\quad a_{i+1} = \lambda _{i}a_{i} - 1. }$$
(2.86)

Note that

$$\displaystyle{ a_{m} = 0\quad \mbox{ and}\quad a_{i} > 0,\quad i \in [0,m - 1]. }$$
(2.87)

Indeed, if a i ≤ 0 for some i ∈ [0, m − 1], then a k < 0 for k ∈ [i + 1, m].

It follows from (2.85) that there exists an n > 0 such that

$$\displaystyle{ \left \vert B^{-n}\tau e_{ 0}\right \vert < 1. }$$
(2.88)

Consider the finite sequence of vectors {w i :   i ∈ [0, m(n + 1)]} defined as follows:

$$\displaystyle{\left \{\begin{array}{lc} w_{i} = a_{i}e_{i}, & \quad i \in [0,m - 1]; \\ w_{m} = B^{-n}\tau e_{0}; & \; \\ w_{m+1+i} = A_{i}w_{m+i},&\quad i \in [0,mn - 1]. \end{array} \right.}$$

Clearly,

$$\displaystyle{w_{km} = B^{k-1-n}\tau e_{ 0},\quad k \in [1,n + 1],}$$

which means that we can consider {w i } as an m(n + 1)-periodic sequence defined for \(i \in \mathbb{Z}\).

Let us note that

$$\displaystyle{A_{i}w_{i} = a_{i}A_{i}e_{i} = a_{i}\frac{v_{i+1}} {\vert v_{i}\vert },\quad i \in [0,m - 2],}$$
$$\displaystyle{w_{i+1} = (\lambda _{i}a_{i} - 1) \frac{v_{i+1}} {\vert v_{i+1}\vert } = a_{i}\frac{v_{i+1}} {\vert v_{i}\vert } - e_{i+1},\quad i \in [0,m - 2],}$$

and

$$\displaystyle{A_{m-1}w_{m-1} = a_{m-1} \frac{v_{m}} {\vert v_{m-1}\vert } = \frac{v_{m}} {\lambda _{m-1}\vert v_{m-1}\vert } = e_{m}}$$

(in the last relation, we take into account that a m−1 λ m−1 = 1 since a m = 0).

The above relations and condition (2.88) imply that

$$\displaystyle{ \vert w_{i+1} - A_{i}w_{i}\vert < 2,\quad i \in \mathbb{Z}. }$$
(2.89)

Now we take a small d > 0 and consider the m(n + 1)-periodic sequence

$$\displaystyle{\xi =\{ x_{i} = p_{i} + dw_{i}:\; i \in \mathbb{Z}\}.}$$

We claim that if d is small enough, then ξ is a 2d-pseudotrajectory of f.

Represent

$$\displaystyle{\,f(x_{i}) = f(\,p_{i}) + Df(\,p_{i})dw_{i} + F_{i}(dw_{i}) = p_{i+1} + A_{i}dw_{i} + F_{i}(dw_{i}),}$$

where F i (v) = o( | v | ) as v → 0.

It follows from estimates (2.77) that

$$\displaystyle{\vert \,f(x_{i}) - x_{i+1}\vert < 2d}$$

for small d.

By Lemma 2.4.3, the m-periodic trajectory { p i } is hyperbolic; hence, { p i } has a neighborhood in which { p i } is the unique periodic trajectory. It follows that if d is small enough, then the pseudotrajectory {x i } is \(2\mathcal{L}d\)-shadowed by { p i }.

The inequalities \(\vert x_{i} - p_{i}\vert \leq 2\mathcal{L}d\) imply that \(\vert a_{i}\vert = \vert w_{i}\vert \leq 2\mathcal{L}\) for 0 ≤ im − 1.

Now the equalities λ i = (a i+1 + 1)∕a i imply that if 0 ≤ im − 1, then

$$\displaystyle{\lambda _{0} \cdot \ldots \cdot \lambda _{i-1} = \frac{a_{1} + 1} {a_{0}} \frac{a_{2} + 1} {a_{1}} \ldots \frac{a_{i} + 1} {a_{i-1}} =}$$
$$\displaystyle{= \frac{a_{i} + 1} {a_{0}} \left (1 + \frac{1} {a_{1}}\right )\ldots \left (1 + \frac{1} {a_{i-1}}\right ) \geq }$$
$$\displaystyle{\geq \frac{1} {2\mathcal{L}}\left (1 + \frac{1} {2\mathcal{L}}\right )^{i-1} > \frac{1} {4\mathcal{L}}\left (1 + \frac{1} {2\mathcal{L}}\right )^{i}}$$

(we take into account that \(1 + 1/(2\mathcal{L}) < 2\) since \(\mathcal{L} \geq 1\)).

It remains to note that

$$\displaystyle{\left \vert Df^{i}(\,p)v_{ u}\right \vert = \lambda _{i-1}\cdots \lambda _{0}\vert v_{u}\vert,\quad 0 \leq i \leq m - 1,}$$

and that we started with an arbitrary vector v u U( p).

This proves our statement for jm − 1. If jm, we take an integer k > 0 such that km > j and repeat the above reasoning for the periodic trajectory p 0, , p km−1 (note that we have not used the condition that m is the minimal period). The lemma is proved. □

In the following lemmas, we return to the case of a diffeomorphism f of a smooth closed manifold M since the reasoning becomes “global.” We still assume that f has the Lipschitz periodic shadowing property and apply analogs of Lemmas 2.4.3 and 2.4.4 for the case of a manifold.

Lemma 2.4.5

The diffeomorphism f satisfies Axiom A.

Proof

Denote by P l the set of points p ∈ Per( f) of index l (as usual, the index of a hyperbolic periodic point is the dimension of its stable manifold).

Let R l be the closure of P l . Clearly, R l is a compact f-invariant set. We claim that any R l is a hyperbolic set. Let n = dimM.

Consider a point qR l and fix a sequence of points p m P l such that p m q as m. By an analog of Lemma 2.4.4, there exist complementary subspaces S( p m ) and U( p m ) of \(T_{p_{m}}M\) (of dimensions l and nl, respectively) for which estimates (H2.1) and (H2.2) hold.

Standard reasoning shows that, introducing local coordinates in a neighborhood of (q, T q M) in the tangent bundle of M, we can select a subsequence \(p_{m_{k}}\) for which the sequences \(S(\,p_{m_{k}})\) and \(U(\,p_{m_{k}})\) converge (in the Grassmann topology) to subspaces of T q M (let S 0 and U 0 be the corresponding limit subspaces).

The limit subspaces S 0 and U 0 are complementary in T q M. Indeed, consider the “angle” \(\beta _{m_{k}}\) between the subspaces \(S(\,p_{m_{k}})\) and \(U(\,p_{m_{k}})\) which is defined (with respect to the introduced local coordinates in a neighborhood of (q, T q M)) as follows:

$$\displaystyle{\beta _{m_{k}} =\min \vert v^{s} - v^{u}\vert,}$$

where the minimum is taken over all possible pairs of unit vectors \(v^{s} \in S(\,p_{m_{k}})\) and \(v^{u} \in U(\,p_{m_{k}})\).

The same reasoning as in the proof of Lemma 2.1.5 shows that the values \(\beta _{m_{k}}\) are estimated from below by a positive constant α = α(N, C, λ). Clearly, this implies that the subspaces S 0 and U 0 are complementary.

It is easy to show that the limit subspaces S 0 and U 0 are unique (which means, of course, that the sequences S( p m ) and U( p m ) converge). For the convenience of the reader, we prove this statement.

To get a contradiction, assume that there is a subsequence \(p_{m_{i}}\) for which the sequences \(S(\,p_{m_{i}})\) and \(U(\,p_{m_{i}})\) converge to complementary subspaces S 1 and U 1 different from S 0 and U 0 (for definiteness, we assume that S 0 ∖S 1 ≠ ∅).

Due to the continuity of Df, the inequalities

$$\displaystyle{\left \vert Df^{j}(q)v\right \vert \leq C\lambda ^{j}\vert v\vert,\quad v \in S_{ 0} \cup S_{1},}$$

and

$$\displaystyle{\left \vert Df^{j}(q)v\right \vert \geq C^{-1}\lambda ^{-j}\vert v\vert,\quad v \in U_{ 0} \cup U_{1},}$$

hold for j ≥ 0.

Since

$$\displaystyle{T_{q}M = S_{0} \oplus U_{0} = S_{1} \oplus U_{1},}$$

our assumption implies that there is a vector vS 0 such that

$$\displaystyle{v = v^{s} + v^{u},\quad v^{s} \in S_{ 1},v^{u} \in U_{ 1},v^{u}\neq 0.}$$

Then

$$\displaystyle{\vert Df^{j}(q)v\vert \leq C\lambda ^{j}\vert v\vert \rightarrow 0,\quad j \rightarrow \infty,}$$

and

$$\displaystyle{\left \vert Df^{j}(q)v\right \vert \geq C^{-1}\lambda ^{-j}\vert v^{u}\vert - C\lambda ^{j}\vert v^{s}\vert \rightarrow \infty,\quad j \rightarrow \infty,}$$

and we get the desired contradiction.

It follows that there are uniquely defined complementary subspaces S(q) and U(q) for qR l with proper hyperbolicity estimates; the Df-invariance of these subspaces is obvious. We have shown that each R l is a hyperbolic set with dimS(q) = l and dimU(q) = nl for qR l .

If rΩ( f), then there exists a sequence of points r m r as m and a sequence of indices k m as m such that \(f^{k_{m}}(r_{m}) \rightarrow r\).

Clearly, if we continue the sequence

$$\displaystyle{r_{m},f(r_{m}),\ldots,f^{k_{m}-1}(r_{ m})}$$

periodically with period k m , we get a periodic d m -pseudotrajectory of f with d m → 0 as m.

Since f has the Lipschitz periodic shadowing property, for large m there exist periodic points p m such that dist( p m , r m ) → 0 as m. Thus, periodic points are dense in Ω( f).

Since hyperbolic sets with different dimensions of the subspaces U(q) are disjoint, we get the equality

$$\displaystyle{\varOmega (\,f) = R_{0} \cup \ldots \cup R_{n},}$$

which implies that Ω( f) is hyperbolic. The lemma is proved. □

Thus, to prove Theorem 2.4.4, it remains to prove the following lemma.

Lemma 2.4.6

If f has the Lipschitz periodic shadowing property, then f satisfies the no cycle condition.

Proof

To simplify presentation, we prove that f has no 1-cycles (in the general case, the idea is literally the same, but the notation is heavy; we leave this case to the reader).

To get a contradiction, assume that

$$\displaystyle{\,p \in (W^{u}(\varOmega _{ i}) \cap W^{s}(\varOmega _{ i}))\setminus \varOmega (\,f).}$$

In this case, there are sequences of indices j m , k m as m such that

$$\displaystyle{\,f^{-j_{m} }(\,p),f^{k_{m} }(\,p) \rightarrow \varOmega _{i},\quad m \rightarrow \infty.}$$

Since the set Ω i is compact, we may assume that

$$\displaystyle{\,f^{-j_{m} }(\,p) \rightarrow q \in \varOmega _{i}\;\mbox{ and}\;f^{k_{m} }(\,p) \rightarrow r \in \varOmega _{i}.}$$

Since Ω i contains a dense positive semitrajectory, there exist points s m r and indices l m > 0 such that \(f^{l_{m}}(s_{m}) \rightarrow q\) as m.

Clearly, if we continue the sequence

$$\displaystyle{\,p,f(\,p),\ldots,f^{k_{m}-1}(\,p),s_{ m},\ldots,f^{l_{m}-1}(s_{ m}),f^{-j_{m} }(\,p),\ldots,f^{-1}(\,p)}$$

periodically with period k m + l m + j m , we get a periodic d m -pseudotrajectory of f with d m → 0 as m.

Since f has the Lipschitz periodic shadowing property, there exist periodic points p m (for m large enough) such that p m p as m, and we get the desired contradiction with the assumption that pΩ( f). The lemma is proved. □

Historical Remarks

Theorem 2.4.1 was published by A. V. Osipov, the first author, and S. B. Tikhomirov in [50].

2.5 Hölder Shadowing for Diffeomorphisms

In this section, we explain the main ideas of the proof of the following result.

Theorem 2.5.1

Assume that a diffeomorphism f of class C 2 of a smooth closed manifold has the Hölder shadowing property on finite intervals with constants \(\mathcal{L},C,d_{0},\theta,\omega\) and that

$$\displaystyle{ \theta \in (1/2,1)\mathit{\mbox{ and }}\theta +\omega > 1. }$$
(2.90)

Then f is structurally stable.

The proof of Theorem 2.5.1 is quite complicated. For that reason, we try to simplify the presentation and omit inessential technical details; the reader can find the original Tikhomirov’s proof in the paper [101].

The main two steps of the proof of Theorem 2.5.1 are as follows.

First one considers a trajectory { p k = f k( p)} of f, denotes A k = Df( p k ), and shows that under conditions of Theorem 2.5.1, the sequence \(\mathcal{A} =\{ A_{k}\}\) has a weak analog of the Perron property (in which the existence of bounded solutions of the inhomogeneous difference equations is replaced by the existence of “slowly growing” solutions).

We reproduce this part of the proof in Theorem 2.5.2 in which we restrict our consideration to the case of a diffeomorphism f of the Euclidean space \(\mathbb{R}^{n}\).

After that, it is shown that the above-mentioned weak analog of the Perron property implies then f satisfies the analytic strong transversality condition (with exponential estimates) and, hence, by the Mañé theorem, f is structurally stable. To explain the basic techniques of that part of the proof, we prove the above statement in Theorem 2.5.3 in the case of a one-dimensional phase space (and note that the reasoning in the proof of Theorem 2.5.3 reproduces the most important part of the proof given by Tikhomirov). We again refer the reader to [101] for the proof of the general case.

Theorem 2.5.2

Assume that a diffeomorphism f of the Euclidean space \(\mathbb{R}^{n}\) has the Hölder shadowing property on finite intervals with constants \(\mathcal{L},C,d_{0},\theta,\omega\) and that condition ( 2.90 ) is satisfied.

Assume, in addition, that there exist constants S, ɛ > 0 such that

$$\displaystyle{ \vert \,f(\,p_{k} + v) - p_{k+1} - A_{k}v\vert \leq S\vert v\vert ^{2},\quad k \in \mathbb{Z},\;\vert v\vert \leq \varepsilon. }$$
(2.91)

Then there exist constants L > 0 and γ ∈ (0, 1) such that for any \(i \in \mathbb{Z}\) and N > 0 and any sequence

$$\displaystyle{ W =\{ w_{k} \in \mathbb{R}^{n}:\; i + 1 \leq k \leq i + N + 1\} }$$
(2.92)

with | w k | ≤ 1, the difference equations

$$\displaystyle{ v_{k+1} = A_{k}v_{k} + w_{k+1},\quad i \leq k \leq i + N, }$$
(2.93)

have a solution

$$\displaystyle{ V =\{ v_{k}:\; i \leq k \leq i + N + 1\}, }$$
(2.94)

such that the value

$$\displaystyle{ \|V \|:=\max _{i\leq k\leq i+N+1}\vert v_{k}\vert }$$
(2.95)

satisfies the estimate

$$\displaystyle{ \|V \| \leq LN^{\gamma }. }$$
(2.96)

Remark 2.5.1

Clearly, an analog of condition (2.91) is satisfied if we consider a diffeomorphism of class C 2 for which the trajectory { p k } is contained in a bounded subset of \(\mathbb{R}^{n}\) (or a diffeomorphism of class C 2 of a smooth closed manifold studied in the original paper [101]). In fact, it was noted by Tikhomirov that one can prove a similar result in the case where exponent 2 in (2.91) is replaced by any ν > 1. The reasoning remains almost the same, but calculations become very cumbersome. For that reason, we follow the proof given in [101] (with exponent 2).

Proof (of Theorem 2.5.2)

Denote

$$\displaystyle{\alpha =\theta -1/2.}$$

Inequalities (2.90) imply that

$$\displaystyle{ \alpha \in (0,1/2)\mbox{ and }1/2 -\alpha <\omega. }$$
(2.97)

Consider two auxiliary linear functions of β ≥ 0,

$$\displaystyle{g_{1}(\beta ) = (2+\beta )(1/2 -\alpha )\mbox{ and }g_{1}(\beta ) = (2+\beta )\omega.}$$

By inequalities (2.97),

$$\displaystyle{g_{2}(0) = 2\omega > 1 - 2\alpha = g_{1}(0) \in (0,1)}$$

and

$$\displaystyle{g_{2}^{{\prime}}(\beta ) =\omega > 1/2 -\alpha = g_{ 1}^{{\prime}}(\beta ).}$$

Hence, there exists a β > 0 such that

$$\displaystyle{g_{1}(\beta ) \in (0,1)\mbox{ and }g_{2}(\beta ) > 1.}$$

We fix such a β and write the above relations in the form

$$\displaystyle{ 0 < (2+\beta )(1/2 -\alpha ) < 1\mbox{ and }(2+\beta )\omega > 1. }$$
(2.98)

Introduce the values

$$\displaystyle{\gamma = ((2+\beta )\omega )^{-1}\mbox{ and }\gamma _{ 1} = 1 - (2+\beta )(1/2 -\alpha ).}$$

Then it follows from (2.98) that

$$\displaystyle{ 0 <\gamma < 1\mbox{ and }\gamma _{1} > 0. }$$
(2.99)

Now we fix a sequence W of the form (2.92) and denote by E(W) the set of all sequences V of the form (2.94) that satisfy Eqs. (2.93). The function ∥V ∥ is continuous on the linear space of sequences V; the set E(W) is closed. Hence, the value

$$\displaystyle{ F(W) =\min _{V \in E(W)}\|V \| }$$
(2.100)

is defined.

The set of finite sequences W of the form (2.92) with

$$\displaystyle{\|W\| =\max _{i+1\leq k\leq N+1}\vert w_{k}\vert \leq 1}$$

is compact. The function F(W) is continuous in W; thus, there exists the number

$$\displaystyle{Q =\max _{W}F(W).}$$

Let us fix sequences W 0 and V 0E(W 0) such that

$$\displaystyle{ Q = F(W_{0}) =\| V _{0}\|. }$$
(2.101)

Note the following two properties of the number Q. They follow from the definition of Q and from the linearity of Eqs. (2.93).

  • (Q1) Any sequence VE(W 0) satisfies the inequality

    $$\displaystyle{\|V \| \geq Q.}$$
  • (Q2) For any sequence W of the form (2.92) there exists a sequence VE(W) such that

    $$\displaystyle{\|V \| \leq Q\|W\|.}$$

It follows from property (Q2) that to complete the proof of our theorem, it is enough to prove the following statement:

There exists a number L independent of i and N such that

$$\displaystyle{ Q \leq LN^{\gamma }. }$$
(2.102)

Define the number

$$\displaystyle{ d =\varepsilon Q^{-(2+\beta )}. }$$
(2.103)

Let us consider the following two cases.

  • Case 1: C((S + 1)d)ω < N. In this case,

    $$\displaystyle{Q < \left (\varepsilon ^{\omega }(S + 1)^{\omega }/C\right )^{\gamma }N^{\gamma },}$$

    which proves inequalities (2.103) with

    $$\displaystyle{L = \left (\varepsilon ^{\omega }(S + 1)^{\omega }/C\right )^{\gamma }.}$$
  • Case 2: C((S + 1)d)ωN. In this case, we prove a stronger statement: There exists a number L independent of i and N such that

    $$\displaystyle{ Q \leq L. }$$
    (2.104)

    Treating Case 2, we assume without loss of generality that i = 0.

Also, without loss of generality, we assume that ɛ < 1 and Q > 2. Concerning the latter assumption, we note that if there exists a fixed number L independent of N such that QL, then estimate (2.104) is obviously valid. Thus, we may assume that Q is larger than any prescribed number independent of N. Applying the same reasoning, we assume that Q is so large that

$$\displaystyle{ Q > ((S + 1)\varepsilon /d_{0})^{1/(2+\beta )} }$$
(2.105)

and

$$\displaystyle{ \mathcal{L}((S + 1)\varepsilon /Q^{2+\beta })^{\theta } <\varepsilon /2. }$$
(2.106)

Fix sequences W 0 and V 0 for which relation (2.101) is valid. To simplify notation, write V 0 = {v k }.

Consider the sequence of points

$$\displaystyle{\,y_{k} = p_{k} + dv_{k},\quad 0 \leq k \leq N + 1.}$$

We claim that this sequence is an (S + 1)d-pseudotrajectory of f.

Let us first note that | v k | ≤ Q; hence,

$$\displaystyle{ \vert dv_{k}\vert \leq \varepsilon Q^{-(2+\beta )}Q =\varepsilon Q^{-(1+\beta )} <\varepsilon /2. }$$
(2.107)

In addition,

$$\displaystyle{ (dQ)^{2} = (\varepsilon Q^{-(1+\beta )})^{2} <\varepsilon Q^{-(2+\beta )} = d. }$$
(2.108)

Now we estimate

$$\displaystyle{\vert \,f(\,y_{k}) - y_{k+1}\vert = \vert \,f(\,p_{k} + dv_{k}) - (\,p_{k+1} + dv_{k+1})\vert =}$$
$$\displaystyle{= \vert \,f(\,p_{k} + dv_{k}) - (\,p_{k+1} + dA_{k}v_{k} + dw_{k+1})\vert \leq }$$
$$\displaystyle{\leq \vert \,f(\,p_{k} + dv_{k}) - (\,p_{k+1} + dA_{k}v_{k})\vert + d\vert w_{k+1}\vert \leq }$$
$$\displaystyle{\leq S\vert dv_{k}\vert ^{2} + d \leq (S + 1)d.}$$

We estimate the first term of the third line taking into account condition (2.91) and inequality (2.107); estimating the first term of the last line, we refer to inequality (2.108).

Inequality (2.105) implies that

$$\displaystyle{Q^{2+\beta } > (S + 1)\varepsilon /d_{ 0};}$$

hence,

$$\displaystyle{(S + 1)d = (S + 1)\varepsilon Q^{-(2+\beta )} < d_{ 0}.}$$

Since we treat Case 2,

$$\displaystyle{N \leq C((S + 1)d)^{-\omega } < Cd^{-\omega },}$$

and we can apply the Hölder shadowing property on finite intervals to conclude that there exists an exact trajectory {x k } of f such that

$$\displaystyle{\vert \,y_{k} - x_{k}\vert \leq \mathcal{L}((S + 1)d)^{\theta },\quad 0 \leq k \leq N + 1.}$$

Denote x k = p k + c k and \(\mathcal{L}_{1} = \mathcal{L}(S + 1)^{\theta }\). Then

$$\displaystyle{ \vert dv_{k} - c_{k}\vert \leq \vert \,y_{k} - x_{k}\vert \leq \mathcal{L}_{1}d^{\theta },\quad 0 \leq k \leq N + 1, }$$
(2.109)

and

$$\displaystyle{ \vert c_{k}\vert \leq Qd + \mathcal{L}_{1}d^{\theta },\quad 0 \leq k \leq N + 1. }$$
(2.110)

Inequalities (2.107) and (2.106) imply that

$$\displaystyle{\vert c_{k}\vert <\varepsilon.}$$

By the first inequality in (2.98),

$$\displaystyle{Q > Q^{(1/2-\alpha )(2+\beta )} = (\varepsilon /d)^{1/2-\alpha } =\varepsilon ^{1/2-\alpha }d^{\alpha -1/2}.}$$

Hence,

$$\displaystyle{Qd >\varepsilon ^{1/2-\alpha }d^{\alpha +1/2} =\varepsilon ^{1/2-\alpha }d^{\theta }.}$$

Now it follows from (2.110) that there exists an \(\mathcal{L}_{2}\) independent of N such that

$$\displaystyle{\vert c_{k}\vert \leq \mathcal{L}_{2}Qd.}$$

Since p k+1 + c k+1 = f( p k + c k ), we can estimate

$$\displaystyle{\vert c_{k+1} - A_{k}c_{k}\vert = \vert \,f(\,p_{k} + c_{k}) - (\,p_{k+1} + A_{k}c_{k}\vert \leq S\vert c_{k})\vert ^{2} \leq S\mathcal{L}_{ 2}(Qd)^{2}.}$$

Denote t k+1 = c k+1A k c k ; then

$$\displaystyle{\vert t_{k}\vert \leq S\vert c_{k}\vert ^{2} \leq \mathcal{L}_{ 3}(Qd)^{2},}$$

where the constant \(\mathcal{L}_{3}\) does not depend on N. By property (Q2), there exists a sequence z k such that

$$\displaystyle{z_{k+1} = A_{k}z_{k} + t_{k+1}\mbox{ and }\vert z_{k}\vert \leq Q\mathcal{L}_{3}(Qd)^{2},\quad 0 \leq k \leq N.}$$

Consider the sequence r k = c k z k . Clearly,

$$\displaystyle{ r_{k+1} = A_{k}r_{k}\mbox{ and }\vert r_{k} - c_{k}\vert \leq Q\mathcal{L}_{3}(Qd)^{2},\quad 0 \leq k \leq N. }$$
(2.111)

Now we define the sequence e k = (dv k r k )∕d. Relations (2.109) and (2.111) imply that

$$\displaystyle{ e_{k+1} = A_{k}e_{k} + w_{k+1},\quad 0 \leq k \leq N, }$$
(2.112)

and

$$\displaystyle{\vert e_{k}\vert = \vert ((dv_{k} - c_{k}) - (r_{k} - c_{k}))/d\vert \leq \mathcal{L}_{1}d^{\theta -1} + \mathcal{L}_{ 3}Q^{3}d,\quad 0 \leq k \leq N.}$$

Property (Q1) implies that

$$\displaystyle{\mathcal{L}_{1}d^{\theta -1} + \mathcal{L}_{ 3}Q^{3}d = \mathcal{L}_{ 1}d^{\alpha -1/2} + \mathcal{L}_{ 3}Q^{3}d \geq Q.}$$

We can apply (2.103) and find \(\mathcal{L}_{4},\mathcal{L}_{5} > 0\) independent of N and such that this inequality takes the form

$$\displaystyle{\mathcal{L}_{4}Q^{-(2+\beta )(\alpha -1/2)} + \mathcal{L}_{ 5}Q^{1-\beta }\geq Q,}$$

or

$$\displaystyle{\mathcal{L}_{4}Q^{1-\gamma _{1} } + \mathcal{L}_{5}Q^{1-\beta }\geq Q.}$$

It follows that either

$$\displaystyle{\mathcal{L}_{4}Q^{1-\gamma _{1} } \geq Q/2}$$

or

$$\displaystyle{\mathcal{L}_{5}Q^{1-\beta }\geq Q/2,}$$

which implies that

$$\displaystyle{Q \leq \max \left ((2\mathcal{L}_{4})^{1/\gamma _{1} },(2\mathcal{L}_{5})^{1/\beta }\right ).}$$

The theorem is proved. □

Now we assume, in addition, that there exists a constant R > 0 such that

$$\displaystyle{ \|A_{k}\| \leq R,\quad k \in \mathbb{Z}. }$$
(2.113)

Remark 2.5.2

Of course, an estimate of the form (2.113) holds for A k = Df( p k ) in the case of a diffeomorphism f of a closed manifold.

Theorem 2.5.3

Let f be a diffeomorphism of the line \(\mathbb{R}\) having the Hölder shadowing property on finite intervals. Assume that conditions ( 2.91 ) and ( 2.113 ) are satisfied for a trajectory { p k = f k( p)}. There exists a constant μ ∈ (0, 1) with the following property.

For any \(k \in \mathbb{Z}\) there exists a constant C > 0 and subspaces S( p k ) and U( p k ) of \(\mathbb{R}\) such that

$$\displaystyle{ S(\,p_{k}) + U(\,p_{k}) = \mathbb{R}, }$$
(2.114)
$$\displaystyle{ \vert A_{k+l-1}\cdots A_{k}v\vert \leq C\mu ^{l}\vert v\vert,\quad v \in S(\,p_{ k}),\;l \geq 0, }$$
(2.115)
$$\displaystyle{ \vert A_{k-l}^{-1}\cdots A_{ k-1}^{-1}v\vert \leq C\mu ^{l}\vert v\vert,\quad v \in U(\,p_{ k}),\;l \geq 0. }$$
(2.116)

The essential part of the proof of Theorem 2.5.3 is contained in the following lemma.

Let us first introduce some notation. Consider a one-dimensional vector (i.e., a real number) e 0 with | e 0 | = 1 and define a sequence \(\{e_{k}:\; k \in \mathbb{Z}\}\) as follows:

$$\displaystyle{ e_{k+1} = A_{k}e_{k}/\vert A_{k}e_{k}\vert,\quad e_{-k-1} = A_{-k-1}^{-1}e_{ -k}/\vert A_{-k-1}^{-1}e_{ -k}\vert,\quad k \geq 0. }$$
(2.117)

Set

$$\displaystyle{\lambda _{k} = \vert A_{k}e_{k}\vert.}$$

It follows from inequalities (2.113) that

$$\displaystyle{ \lambda _{k} \in [1/R,R],\quad k \in \mathbb{Z}. }$$
(2.118)

Set also

$$\displaystyle{ \varPi (k,l) = \lambda _{k}\cdots \lambda _{k+l-1},\quad k \in \mathbb{Z},\;l \geq 1. }$$
(2.119)

Lemma 2.5.1

If the sequence \(\mathcal{A}\) satisfies the conclusion of Theorem  2.5.2 , then there exists a number N depending only on L, γ, and R (see inequality ( 2.113 )) and such that, for any \(i \in \mathbb{Z}\) , one of the following alternatives is valid:

$$\displaystyle{ \mathit{\mbox{ either }}\varPi (i,N) > 2\mathit{\mbox{ or }}\varPi (i + N,N) < 1/2. }$$
(2.120)

Proof

Fix numbers \(i \in \mathbb{Z}\) and N > 0 and consider the sequence

$$\displaystyle{w_{k} = -e_{k},\quad i \leq k \leq i + 2N + 1.}$$

It follows from the conclusion of Theorem 2.5.2 that there exists a sequence

$$\displaystyle{\{v_{k}:\; i \leq k \leq i + 2N\}}$$

such that

$$\displaystyle{v_{k+1} = A_{k}v_{k} + w_{k+1}\mbox{ and }\vert v_{k}\vert \leq L(2N + 1)^{\gamma },\quad i \leq k \leq i + 2N.}$$

Set v k = a k e k , where \(a_{k} \in \mathbb{R}\). Then

$$\displaystyle{ a_{k+1} = \lambda _{k}a_{k} - 1\mbox{ and }\vert a_{k}\vert \leq L(2N + 1)^{\gamma },\quad i \leq k \leq i + 2N. }$$
(2.121)

Now we show that there exists a large enough number N (depending only on L, γ, and R) such that if a i+N ≥ 0, then Π(i, N) > 2, and if a i+N < 0, then Π(i + N, N) < 1∕2.

Let us prove the existence of N for the first case (i.e., for the case where a i+N ≥ 0).

Since λ k > 0, it follows from relations (2.121) that if a k ≤ 0 for some k ∈ [i, i + 2N − 1], then a k+1 < 0. Thus, if a i+N ≥ 0, then a i , , a i+N−1 > 0.

Relations (2.121) imply that in this case,

$$\displaystyle{\lambda _{k} = \frac{a_{k+1} + 1} {a_{k}},\quad i \leq k \leq i + N - 1.}$$

Hence,

$$\displaystyle{\varPi (i,N) = \frac{a_{i+1} + 1} {a_{i}} \frac{a_{i+2} + 1} {a_{i+1}} \cdots \frac{a_{i+N} + 1} {a_{i+N-1}} =}$$
$$\displaystyle{= \frac{1} {a_{i}} \frac{a_{i+1} + 1} {a_{i+1}} \frac{a_{i+2} + 1} {a_{i+2}} \cdots \frac{a_{i+N-1} + 1} {a_{i+N-1}} (a_{i+N} + 1) =}$$
$$\displaystyle{= \frac{a_{i+N} + 1} {a_{i}} \prod _{k=i+1}^{i+N-1}\frac{a_{k} + 1} {a_{k}},}$$

and it follows from relations (2.121) that

$$\displaystyle{ \varPi (i,N) \geq \frac{1} {L(2N + 1)^{\gamma }}\left (1 + \frac{1} {L(2N + 1)^{\gamma }}\right )^{N-1}. }$$
(2.122)

Denote the expression on the right in (2.122) by G 1(γ, N). Since

$$\displaystyle{\log G_{1}(\gamma,N) = -\gamma \log (L(2N + 1)) + (N - 1)\log \left (1 + \frac{1} {L(2N + 1)^{\gamma }}\right ),}$$
$$\displaystyle{\log \left (1 + \frac{1} {L(2N + 1)^{\gamma }}\right ) \simeq (L(2N + 1))^{-\gamma }}$$

for large N, and γ ∈ (0, 1), we conclude that

$$\displaystyle{G_{1}(\gamma,N) \rightarrow \infty,\quad N \rightarrow \infty.}$$

Hence, there exists an N 1 depending only on L and γ such that G 1(γ, N) > 2 for NN 1.

Now we consider the second case, i.e., we assume that a i+N < 0. In this case, it follows from relations (2.121) that

$$\displaystyle{ a_{k} \in (-L(2N + 1)^{\gamma },-1),\quad i + N < k \leq i + 2N. }$$
(2.123)

As above, we set

$$\displaystyle{\lambda _{k} = \frac{a_{k+1} + 1} {a_{k}}.}$$

Now we write

$$\displaystyle{\varPi (i + N + 1,N - 1) = \frac{a_{i+N+2} + 1} {a_{i+N+1}} \frac{a_{i+N+3} + 1} {a_{i+N+2}} \cdots \frac{a_{i+2N} + 1} {a_{i+2N-1}} =}$$
$$\displaystyle{= \frac{1} {a_{i+N+1}} \frac{a_{i+N+2} + 1} {a_{i+N+2}} \frac{a_{i+N+3} + 1} {a_{i+N+3}} \cdots \frac{a_{i+2N-1} + 1} {a_{i+2N-1}} (a_{i+2N} + 1)}$$

and conclude that

$$\displaystyle{ \varPi (i + N + 1,N - 1) = \frac{a_{i+2N} + 1} {a_{i+N+1}} \prod _{k=i+N+2}^{i+2N-1}\frac{a_{k} + 1} {a_{k}}. }$$
(2.124)

Inclusions (2.123) imply that

$$\displaystyle{0 < \frac{a_{k} + 1} {a_{k}} < 1 - \frac{1} {L(2N + 1)^{\gamma }},\quad i + N + 2 \leq k \leq i + 2N - 1,}$$

and

$$\displaystyle{0 < \frac{a_{i+2N} + 1} {a_{i+N+1}} < L(2N + 1)^{\gamma }.}$$

Combining these inequalities with (2.124), we conclude that

$$\displaystyle{\varPi (i + N + 1,N - 1) < L(2N + 1)^{\gamma }\left (1 - \frac{1} {L(2N + 1)^{\gamma }}\right )^{N-2}.}$$

Denote the right-hand side of the above inequality by G 2(γ, N). Clearly, G 2(γ, N) → 0 as N; hence, there exists an N 2 depending only on L, γ, and R such that

$$\displaystyle{G(\gamma,N) < \frac{1} {2R},\quad N \geq N_{2}.}$$

If NN 2, then

$$\displaystyle{\varPi (i + N,N) =\lambda _{i+N}\varPi (i + N + 1,N - 1) < R \frac{1} {2R} = 1/2.}$$

Hence, the conclusion of our lemma holds for N = max(N 1, N 2). □

Proof (of Theorem 2.5.3)

Take an arbitrary \(i \in \mathbb{Z}\) and the number N given by Lemma 2.5.1. The following statements hold:

  1. (a)

    If Π(i, N) > 2, then Π(iN, N) > 2;

  2. (b)

    If Π(i, N) < 1∕2, then Π(i + N, N) < 1∕2.

Let us prove statement (a); the proof of statement (b) is similar.

By Lemma 2.5.1 applied to iN, either Π(iN, N) > 2 or Π(i, N) < 1∕2. By the assumption of statement (a), the second case is impossible; thus, Π(iN, N) > 2.

It follows from these statements that only one of the following cases is realized:

  • Case 1. Π(i, N) > 2 for all \(i \in \mathbb{Z}\).

  • Case 2. Π(i, N) < 1∕2 for all \(i \in \mathbb{Z}\).

  • Case 3. There exist indices \(i,j \in \mathbb{Z}\) such that Π(i, N) > 2 and Π( j, N) < 1∕2.

Now we show that Theorem 2.5.3 is valid with μ = 2−1∕N.

Consider Case 1. Take e 0 with | e 0 | = 1 and define \(e_{k},\;k \in \mathbb{Z}\), by formulas (2.117). Represent any integer l ≥ 0 in the form

$$\displaystyle{l = nN + l_{1},\quad n \in \mathbb{Z}_{+},\;0 \leq l_{1} < N.}$$

Then

$$\displaystyle{\varPi (i,l) =\varPi (i,nN)\varPi (i + nN,l_{1}) > 2^{n}R^{-l_{1} }}$$

(in the last estimate, we take into account inequalities (2.118)).

Hence, in Case 1,

$$\displaystyle{ \varPi (i,l) > R^{-l_{1} }\left (2^{-l_{1}/N}\right )\left (2^{1/N}\right )^{l} > C_{ 0}\mu ^{-l},\quad i \in \mathbb{Z},\;l \geq 0, }$$
(2.125)

where

$$\displaystyle{C_{0} = R^{-N}/2.}$$

Now we fix a point p k of the trajectory { p k } and set S( p k ) = {0} and \(U(\,p_{k})=\mathbb{R}\). Clearly, in this case, relations (2.114) and (2.115) are satisfied. Let us prove inequalities (2.116). Take any \(v \in \mathbb{R} = U(\,p_{k})\) and l ≥ 0. Let

$$\displaystyle{w = A_{k-l}^{-1}\cdots A_{ k-1}^{-1}v.}$$

Then

$$\displaystyle{v = A_{k-1}\cdots A_{k-l}w.}$$

Hence,

$$\displaystyle{\vert v\vert = \lambda _{k-l}\cdots \lambda _{k-1}\vert w\vert =\varPi (k - l,l)\vert w\vert,}$$

and it follows from (2.125) that

$$\displaystyle{\vert w\vert \leq C\mu ^{l}\vert v\vert,}$$

where C = (C 0)−1, as required.

In Case 2, we set U( p k ) = {0} and \(S(\,p_{k}) = \mathbb{R}\) and apply a similar reasoning.

Let us now consider Case 3. By our remark at the beginning of the proof,

$$\displaystyle{\varPi (i - nN,N) > 2\mbox{ and }\varPi (\,j + nN,N) < 1/2,\quad n \in \mathbb{Z}_{+}.}$$

In this case, we set \(S(\,p_{k}) = U(\,p_{k}) = \mathbb{R}\). Clearly, in this case, relation (2.114) is satisfied. Let us show how to prove inequalities (2.115).

We treat in detail two cases:

  • Case (I). k + lj

    and

  • Case (II). k < j and k + l > j

    (the remaining cases and the proof of inequalities (2.116) are left to the reader).

In Case (I), we note that ljk and estimate

$$\displaystyle{\varPi (k,l) \leq R^{j-k} = R^{j-k}2^{l/N}2^{-l/N} \leq C\mu ^{l},}$$

where C = R jk2( jk)∕N. Hence,

$$\displaystyle{\vert A_{k+l-1}\cdots A_{k}v\vert \leq C\mu ^{l}\vert v\vert,\quad v \in S(\,p_{ k}).}$$

In Case (II), we represent k + l = j + nN + l 1, where \(n \in \mathbb{Z}_{+}\) and 0 ≤ l 1 < N. Then

$$\displaystyle{\varPi (k,l) =\varPi (k,j - k)\varPi (\,j,nN)\varPi (\,j + nN,l_{1}).}$$

We note that Π(k, jk) ≤ R jk,

$$\displaystyle{\varPi (\,j,nN) < 2^{-n} = 2^{l_{1}/N}\mu ^{l},}$$

and

$$\displaystyle{\varPi (\,j + nN,l_{1}) \leq R^{l_{1} } < R^{N},}$$

which gives us the desired estimate Π(k, l) < l (and, hence, inequalities (2.115)) with C = 2R jk+N. □

Historical Remarks

Theorem 2.5.1 was published by S. B. Tikhomirov in [101].

Let us mention that earlier S. M. Hammel, J. A. Yorke, and C. Grebogi, based on results of numerical experiments, conjectured that a generic dissipative mapping \(f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) belongs to a class \(\mbox{ FHSP}_{D}(\mathcal{L},C,d_{0},1/2,1/2)\) [23, 24]. If this conjecture is true, then, in a sense, Theorem 2.5.1 cannot be improved.

2.6 A Homeomorphism with Lipschitz Shadowing and a Nonisolated Fixed Point

Consider the segment

$$\displaystyle{I_{0} = [-7/6,4/3]}$$

and a mapping f 0:   I 0I 0 defined as follows:

$$\displaystyle{\,f_{0}(x) = \left \{\begin{array}{ll} 1 + (x - 1)/2, &x \in [1/3,4/3]; \\ 2x, &x \in (-1/3,1/3); \\ - 1 + (x + 1)/2,&x \in [-7/6,-1/3].\\ \end{array} \right.}$$

Clearly, the restriction f of f 0 to [−1, 1] is a homeomorphism of [−1, 1] having three fixed points: the points x = ±1 are attracting and the point x = 0 is repelling (and this homeomorphism f is an example of the so-called “North Pole – South Pole” dynamical system; every trajectory starting at a point x ≠ 0, ±1 tends to an attractive fixed point as time tends to + and to the repelling fixed point as time tends to −).

Now we define a homeomorphism f:   [−1, 1] → [−1, 1]. For an integer n ≥ 0, denote \(\mathcal{N}_{n} = 2^{-(n+2)}\) and set

$$\displaystyle{ f(x) =\mathcal{ N}_{n}f_{0}(\mathcal{N}_{n}^{-1}(x - 3\mathcal{N}_{ n})) + 3\mathcal{N}_{n},\quad x \in (2\mathcal{N}_{n},4\mathcal{N}_{n}]. }$$
(2.126)

This defines f on (0, 1]. Set f(0) = 0 and f(x) = −f(−x) for x ∈ [−1, 0).

Clearly, f is a homeomorphism with a nonisolated fixed point x = 0 (for example, every point x = ±2n is fixed). Let us note that in a neighborhood of any fixed point (with the exception of x = 0), f is either linearly expanding with coefficient 2 or linearly contracting with coefficient 1/2.

Theorem 2.6.1

The homeomorphism f has the Lipschitz shadowing property.

Before proving Theorem 2.6.1, we prove two auxiliary lemmas.

Lemma 2.6.1

The mapping f 0 has the Lipschitz shadowing property on I 0 .

Proof

Let

$$\displaystyle{G_{0} = (-1/3,1/3)}$$

and

$$\displaystyle{G_{1} = (-7/6,-1/3) \cup (1/3,4/3).}$$

We take d 0 small enough and dd 0; in fact, we write below several explicit conditions on d and assume that they are satisfied.

There exist trivial cases where ξ is a subset of one of the segments J 1 = [−7∕6, −1∕3], J 2 = [−1∕3, 1∕3], or J 3 = [1∕3, 4∕3].

Let, for example, ξJ 3. The inequalities 1∕3 ≤ x k ≤ 4∕3 imply that

$$\displaystyle{1/2 < 2/3 - d \leq f_{0}(x_{k}) - d \leq x_{k+1} \leq f_{0}(x_{k}) + d \leq 7/6 + d < 15/12.}$$

These inequalities are satisfied for an arbitrary k; hence, ξ belongs to a domain in which f 0 is a hyperbolic diffeomorphism (and ξ is uniformly separated from the boundaries of the domain); by Theorem 1.4.2 (which, of course, is valid for infinite pseudotrajectories as well), there exist \(\mathcal{L},d_{0} > 0\) such that if dd 0, then ξ is \(\mathcal{L}d\)-shadowed by an exact trajectory of f 0.

A similar reasoning can be applied if ξJ 1 or ξJ 2.

To consider “nontrivial” cases, let us first describe possible positions of d-pseudotrajectories ξ of f 0 with small d with respect to J 1, , J 3.

First we show that such a pseudotrajectory cannot intersect both J 1 and J 3. Indeed, if we assume that ξJ 3 ≠ ∅, i.e., there exists an index m such that x m ≥ 1∕3, then

$$\displaystyle{x_{m+1} \geq f_{0}(1/3) - d = 2/3 - d > 1/3}$$

and, consequently,

$$\displaystyle{x_{m+i} > 1/3,\quad i > 0.}$$

Similarly, if there exists an index l such that x l ≤ −1∕3, then

$$\displaystyle{x_{l+1} \leq -2/3 + d < -1/3}$$

and

$$\displaystyle{x_{l+i} < -1/3,\quad i > 0,}$$

and we get a contradiction.

Thus, it remains to consider the cases where either

$$\displaystyle{\xi \subset J_{2} \cup J_{3},\quad \xi \cap \mbox{ Int}(\,J_{2})\neq \emptyset,\quad \xi \cap \mbox{ Int}(\,J_{3})\neq \emptyset,}$$

or

$$\displaystyle{\xi \subset J_{1} \cup J_{2},\quad \xi \cap \mbox{ Int}(\,J_{1})\neq \emptyset,\quad \xi \cap \mbox{ Int}(\,J_{2})\neq \emptyset.}$$

We consider the first case; the reasoning in the second case is similar.

We claim that in the case considered, ξ contains two points x k , x l such that

$$\displaystyle{ 0 < x_{k} < 1/3 < x_{l}. }$$
(2.127)

The existence of the point x l follows directly from our assumption; it is easily seen that

$$\displaystyle{ x_{l+i} \geq 2/3 - d > 1/2,\quad i > 0. }$$
(2.128)

Thus, either the set

$$\displaystyle{\left \{m:\, x_{m} \in \mbox{ Int}(\,J_{2}),x_{m} \leq 0\right \}}$$

is empty (which implies that there exists an index k for which inequality (2.127) is valid) or it is nonempty and bounded from above. In the latter case, let m 0 be its maximal element. Then

$$\displaystyle{x_{m_{0}+1} \leq f_{0}(x_{m_{0}}) + d \leq d}$$

(i.e., \(x_{m_{0}+1} \in J_{2}\)) and \(x_{m_{0}+1} > 0\); thus, we get the required k = m 0 + 1.

Obviously, l > k (see (2.128)). Consider the finite set of indices

$$\displaystyle{\kappa =\{\, i \in [k,l - 1]:\; x_{i} \leq 1/3\}.}$$

This set is nonempty (kκ) and finite; hence, it contains the maximal element. Let it be \(x_{k_{0}}\); clearly,

$$\displaystyle{x_{k_{0}} \leq 1/3 < x_{k_{0}+1}.}$$

To simplify notation, let us assume that k 0 = 0. Thus,

$$\displaystyle{x_{0} \leq 1/3 < x_{1}.}$$

In this case,

$$\displaystyle{ x_{i} \geq 2/3 - d > 1/2,\quad i \geq 2. }$$
(2.129)

On the other hand,

$$\displaystyle{x_{1} \leq 2/3 + d < 1,}$$

and one easily shows that

$$\displaystyle{ x_{i} \leq 1 + 2d,\quad i \geq 2. }$$
(2.130)

Since f 0 −1 has Lipschitz constant 2, ξ is a 2d-pseudotrajectory of f 0 −1; hence,

$$\displaystyle{x_{-1} \leq 1/6 + 2d < 2/9,}$$

and, applying the same reasoning as above, we conclude that

$$\displaystyle{ -4d < x_{i} < 1/6 + 2d < 2/9,\quad i < 0. }$$
(2.131)

Now we show that there exists a d 0 such that if dd 0 and p = x 0, then

$$\displaystyle{ \left \vert \,f_{0}^{k}(\,p) - x_{ k}\right \vert < 3d,\quad k \in \mathbb{Z}. }$$
(2.132)

First, clearly,

$$\displaystyle{\vert \,f_{0}(\,p) - x_{1}\vert < d.}$$

Since the Lipschitz constant of f 0 is 2,

$$\displaystyle{\left \vert \,f_{0}^{2}(\,p) - x_{ 2}\right \vert \leq \left \vert \,f(\,f(\,p)) - f(x_{1})\right \vert + \left \vert \,f(x_{1}) - x_{2}\right \vert < 2d + d = 3d.}$$

It follows from (2.129) that

$$\displaystyle{\,f_{0}^{2}(\,p) > 1/2 - 3d > 1/3,}$$

and then

$$\displaystyle{\,f_{0}^{k}(\,p) > 1/3,\quad k \geq 2.}$$

Hence,

$$\displaystyle{\left \vert \,f_{0}^{3}(\,p) - x_{ 3}\right \vert \leq \left \vert \,f_{0}(\,f_{0}^{2}(\,p)) - f_{ 0}(x_{2})\right \vert + \left \vert \,f_{0}(x_{2}) - x_{3}\right \vert < 3d/2 + d < 3d.}$$

Repeating these estimates, we establish inequalities (2.132) for k ≥ 2.

On the other hand, the inclusion pJ 2 implies that f 0 k( p) ∈ J 2 for k ≤ 0. Since f 0 −1(x) = x∕2 for xJ 2 and (2.131) holds, the inequality

$$\displaystyle{\vert \,f_{0}(x_{1}) - p\vert < d}$$

implies that

$$\displaystyle{\vert x_{1} - f_{0}^{-1}(\,p)\vert < d/2.}$$

After that, we estimate

$$\displaystyle{\left \vert x_{2} - f_{0}^{-2}(\,p)\right \vert \leq \left \vert x_{ 2} - f_{0}^{-1}(x_{ -1})\right \vert +\left \vert \,f_{0}^{-1}(x_{ -1}) - f_{0}^{-1}(\,f_{ 0}^{-1}(\,p))\right \vert < d/2+d/2,}$$

and so on, which shows that an analog of (2.132) with 3d replaced by d holds for k < 0. □

The following statement is almost obvious.

Lemma 2.6.2

Let g be a mapping of a segment J and let numbers M > 0 and m be given. Consider the mapping

$$\displaystyle{g'(\,y) = M^{-1}g(M(\,y - m)) + m}$$

on the set

$$\displaystyle{\,J' =\{ y:\; M(\,y - m) \in J\}.}$$

If g has the Lipschitz shadowing property with constants \(\mathcal{L},d_{0}\) , then g′ has the Lipschitz shadowing property with constants \(\mathcal{L},M^{-1}d_{0}\) .

Proof

First we note that if {y k } is a d-pseudotrajectory of g′ with dd 0M and x k = M( y k m), then

$$\displaystyle{g(x_{k}) - x_{k+1} = M(g'(\,y_{k}) - y_{k+1}).}$$

Hence, {x k } is an Md-pseudotrajectory of g.

Since Mdd 0, there exists a point p such that

$$\displaystyle{\left \vert g^{k}(\,p) - x_{ k}\right \vert \leq \mathcal{L}Md.}$$

Set p′ = M −1 p + m. Then, obviously,

$$\displaystyle{\left \vert (g')^{k}(\,p') - y_{ k}\right \vert = M^{-1}\left \vert g^{k}(\,p) - x_{ k}\right \vert \leq \mathcal{L}d.}$$

Let us prove Theorem 2.6.1.

Proof

For a natural n, define the segment

$$\displaystyle{I_{n} = [\alpha _{n},\beta _{n}] = [11\mathcal{N}_{n}/6,13\mathcal{N}_{n}/3]}$$

and note that formula (2.126) defining f for \(x \in (2\mathcal{N}_{n},4\mathcal{N}_{n}]\) is, in fact, valid for xI n .

To prove Theorem 2.6.1, we first claim that there exists a constant c independent of n such that if d satisfies a condition of the form

$$\displaystyle{ d \leq c\mathcal{N}_{n} }$$
(2.133)

and ξ = {x k } is a d-pseudotrajectory of f that intersects I n , then ξ is a subset of one of the segments I n−1, I n , I n+1.

In fact, all the conditions imposed below on d have the form (2.133).

It follows from the inequalities

$$\displaystyle{\,f(\alpha _{n}) = 23\mathcal{N}_{n}/12 > \alpha _{n},\quad f(\beta _{n}) = 25\mathcal{N}_{n}/6 <\beta _{n}}$$

that if c is small enough (we do not repeat this assumption below), then

$$\displaystyle{ \mbox{ Cl}(N(d,f(I_{m}))) \subset I_{m},\quad m = n - 1,n,n + 1. }$$
(2.134)

Thus, if x k I m for some m = n − 1, n, n + 1, then it follows from (2.134) that

$$\displaystyle{ x_{k+i} \in I_{m},\quad i \geq 0. }$$
(2.135)

Let x 0I n .

We assume that

$$\displaystyle{\mbox{ Cl}(N(2d,f^{-1}(I_{ n}))) \subset I_{n-1} \cup I_{n} \cup I_{n+1}}$$

(note that this condition on d has precisely form (2.133)).

By (2.135), x k I n for k ≥ 0. Thus, if the inclusion ξI n does not hold, there exists an index l < 0 such that

$$\displaystyle{x_{l} \in (I_{n-1} \cup I_{n+1})\setminus I_{n}}$$

(recall that ξ is a 2d-pseudotrajectory of f −1).

Assume, for definiteness, that x l I n−1 (the remaining case is treated similarly). In this case, the same inclusions (2.135) imply that

$$\displaystyle{x_{l+i} \in I_{n-1},\quad i \geq 0.}$$

To show that

$$\displaystyle{x_{l+i} \in I_{n-1},\quad i < 0,}$$

take an index m < l and assume that x m I ν . Then inclusions (2.135) imply that

$$\displaystyle{x_{0},x_{l} \in I_{\nu };}$$

hence,

$$\displaystyle{I_{\nu } \cap I_{n}\neq \emptyset \mbox{ and }I_{\nu } \cap I_{n-1}\neq \emptyset,}$$

from which it follows that either ν = n or ν = n − 1. But since x l I n , νn, and we conclude that ξI n−1, as claimed.

Of course, a similar statement holds for the segments I n = [−β n , −α n ].

Without loss of generality, we assume that

$$\displaystyle{ c \leq d_{0}/2, }$$
(2.136)

where d 0 is given by Lemma 2.6.1. Let \(\delta (m) = c\mathcal{N}_{m}\).

Consider a d-pseudotrajectory ξ = {x k } ⊂ [−1, 1] of f with dd 0. If

$$\displaystyle{d \geq \delta (0) = c\mathcal{N}_{0} = c/4,}$$

then 1 ≤ 4dc, and ξ is 4dc-shadowed by the fixed point x = 0.

Otherwise, we find the maximal index m 0 for which d < δ(m 0). In this case,

$$\displaystyle{ d \geq \delta (m_{0} + 1) = \delta (m_{0})/2. }$$
(2.137)

First we assume that

$$\displaystyle{ \xi \cap I_{m}\neq \emptyset \mbox{ for some }m \leq m_{0} }$$
(2.138)

(the case of I m is similar).

In this case, the inequalities

$$\displaystyle{d < \delta (m_{0}) \leq \delta (m)}$$

imply that ξ is a subset of one of the segments I m−1, I m , I m+1. We assume that ξI m+1; in the remaining cases, the same estimates work.

Since

$$\displaystyle{d \leq \delta (m) = c\mathcal{N}_{m} \leq d_{0}\mathcal{N}_{m}/2 = d_{0}\mathcal{N}_{m+1}}$$

(we refer to (2.136)), Lemma 2.6.2 implies that ξ is \(\mathcal{L}\)-shadowed.

If relation (2.138) does not hold, then

$$\displaystyle{\vert x_{k}\vert \leq \alpha _{m_{0}} = \frac{11N_{m_{0}}} {6} = \frac{11\delta (m_{0})} {6c} \leq \frac{11} {3c}d}$$

(we take into account inequality (2.137) in the last estimate). Thus, in this case, ξ is 11d∕(3c)-shadowed by the fixed point x = 0. □

Historical Remarks

In this section, we give a simplified proof of Theorem 2.6.1 compared to the original variant published by A. A. Petrov and the first author in [59].

2.7 Lipschitz Shadowing Implies Structural Stability: The Case of a Vector Field

Let M be a smooth closed manifold with Riemannian metric dist and let X be a vector field on M of class C 1. Denote by ϕ(t, x) the flow on M generated by the vector field X.

Our main goal in this section is to prove the following statement.

Theorem 2.7.1

If a vector field X has the Lipschitz shadowing property, then X is structurally stable.

In the proof of Theorem 2.7.1, we refer to Theorem 1.3.14.

Define a diffeomorphism f on M by setting f(x) = ϕ(1, x).

It is an easy exercise to show that the chain recurrent set \(\mathcal{R}(\phi )\) of the flow ϕ (see Definition 1.3.22) coincides with the chain recurrent set of the diffeomorphism f.

2.7.1 Discrete Lipschitz Shadowing for Flows

In this section, we introduce the notion of discrete Lipschitz shadowing for a vector field in terms of the diffeomorphism f(x) = ϕ(1, x) introduced above and show that the Lipschitz shadowing property of ϕ implies the discrete Lipschitz shadowing.

Definition 2.7.1

A vector field X has the discrete Lipschitz shadowing property if there exist d 0, L > 0 such that if y k M is a sequence with

$$\displaystyle{ \mbox{ dist}(\,y_{k+1},f(\,y_{k})) \leq d \leq d_{0},\quad k \in \mathbb{Z}, }$$
(2.139)

then there exist sequences x k M and \(t_{k} \in \mathbb{R}\) such that

$$\displaystyle{ \vert t_{k} - 1\vert \leq Ld,\;\mbox{ dist}(x_{k},y_{k}) \leq Ld,\;x_{k+1} =\phi (t_{k},x_{k}),\quad k \in \mathbb{Z}. }$$
(2.140)

Lemma 2.7.1

The Lipschitz shadowing property of ϕ implies the discrete Lipschitz shadowing of X.

Proof

First we note that since M is compact and X is C 1-smooth, there exists a ν > 0 such that

$$\displaystyle{ \mbox{ dist}(\phi (t,x),\phi (t,y)) \leq \nu \mbox{ dist}(x,y),\quad x,y \in M,\;t \in [0,1]. }$$
(2.141)

Consider a sequence y k that satisfies inequalities (2.139) and define a mapping \(y:\; \mathbb{R} \rightarrow M\) by setting

$$\displaystyle{\,y(t) =\phi (t - k,y_{k}),\quad k \leq t < k + 1,\;k \in \mathbb{Z}.}$$

Fix a τ ∈ [k, k + 1). If t ∈ [0, 1] and τ + t < k + 1, then

$$\displaystyle{\mbox{ dist}\left (\,y(\tau +t),\phi (t,y(\tau ))\right ) = \mbox{ dist}\left (\phi (\tau +t - k,y_{k}),\phi (t,\phi (\tau -k,y_{k}))\right ) = 0.}$$

If k + 1 ≤ τ + t, then

$$\displaystyle{\mbox{ dist}(\,y(\tau +t),\phi (t,y(\tau ))) = \mbox{ dist}(\phi (\tau +t - k - 1,y_{k+1}),\phi (\tau +t - k,y_{k})) =}$$
$$\displaystyle{= \mbox{ dist}(\phi (\tau +t - k - 1,y_{k+1}),\phi (\tau +t - k - 1,\phi (1,y_{k}))) \leq \nu d.}$$

Thus, y(t) is a (ν + 1)d-pseudotrajectory of ϕ. Hence, if dd 0∕(ν + 1), where d 0 is from the definition of the Lipschitz shadowing property for ϕ, then there exists a trajectory x(t) of X and a reparametrization

$$\displaystyle{\alpha \in \mbox{ Rep}(\mathcal{L}(\nu +1)d)}$$

such that

$$\displaystyle{\mbox{ dist}(\,y(t),x(\alpha (t))) \leq \mathcal{L}(\nu +1)d,\quad t \in \mathbb{R}.}$$

If we set

$$\displaystyle{x_{k} = x(\alpha (k))\mbox{ and }t_{k} = \alpha (k + 1) -\alpha (k),}$$

then

$$\displaystyle{x_{k+1} = x(\alpha (k + 1)) =\phi (\alpha (k + 1) -\alpha (k),x(\alpha (k)) =\phi (t_{k},x_{k}),}$$
$$\displaystyle{\mbox{ dist}(x_{k},y_{k}) = \mbox{ dist}(x(\alpha (k)),y_{k}) \leq \mathcal{L}(\nu +1)d,}$$

and

$$\displaystyle{\vert t_{k} - 1\vert = \left \vert \frac{\alpha (k + 1) -\alpha (k)} {k + 1 - k} - 1\right \vert \leq \mathcal{L}(\nu +1)d.}$$

Taking \(L = \mathcal{L}(\nu +1)\) and d 0 in Definition 2.7.1 as d 0∕(ν + 1), we complete the proof of the lemma. □

As in Sect. 2.3, we reduce our shadowing problem to the problem of existence of bounded solutions of certain difference equations. To clarify the presentation, we again first take \(M = \mathbb{R}^{n}\), assume that the considered vector field X defines a flow (every trajectory is defined for \(t \in \mathbb{R}\)), and assume that the diffeomorphism f satisfies Condition S formulated in Sect. 2.3 (see estimate (2.52)). To treat the general case of a compact manifold M, one has to apply exponential mappings (see Remark 2.7.1 below); we leave details to the reader.

As above, we denote

$$\displaystyle{\|V \| =\sup _{k\in \mathbb{Z}}\vert v_{k}\vert }$$

for a bounded sequence of vectors \(V =\{ v_{k}:\; k \in \mathbb{Z}\}\).

Lemma 2.7.2

Assume that X has the discrete Lipschitz shadowing property with constant L. Let x(t) be an arbitrary trajectory of X, let p k = x(k), and set A k = Df( p k ) (recall that f(x) = ϕ(1, x)). Assume that f satisfies Condition S formulated in Sect.  2.3 . Let \(B =\{ b_{k} \in \mathbb{R}^{n}\}\) be a bounded sequence and denote β 0 = ∥B.

Then there exists a sequence of scalars s k with

$$\displaystyle{\vert s_{k}\vert \leq \beta = L(\beta _{0} + 1)}$$

such that the difference equation

$$\displaystyle{ v_{k+1} = A_{k}v_{k} + X(\,p_{k+1})s_{k} + b_{k+1} }$$
(2.142)

has a solution V = {v k } with

$$\displaystyle{ \|V \| \leq \beta. }$$
(2.143)

Proof

Fix a natural number N and define \(\varDelta _{k} \in \mathbb{R}^{n}\) as the solution of

$$\displaystyle{v_{k+1} = A_{k}v_{k} + b_{k+1},\quad k = -N,\ldots,N - 1,}$$

with Δ N = 0. Then

$$\displaystyle{ \vert \varDelta _{k}\vert \leq C,\quad k = -N,\ldots,N, }$$
(2.144)

where C depends on N, β 0, and an upper bound of ∥A k ∥ for k = −N, , N − 1.

Fix a small number d > 0 and fix μ in (2.52) so that

$$\displaystyle{ \mu C < 1. }$$
(2.145)

Consider the sequence of points \(y_{k} \in \mathbb{R}^{n}\) defined as follows: y k = p k for k ≤ −N,

$$\displaystyle{\,y_{k} = p_{k} + d\varDelta _{k},\quad k = -N,\ldots,N - 1,}$$

and y N+k = f k( y N ) for k > 0.

Then y k+1 = f( y k ) for k ≤ −N − 1 and kN.

Since

$$\displaystyle{\,y_{k+1} = p_{k+1} + d\varDelta _{k+1} = p_{k+1} + dA_{k}\varDelta _{k} + db_{k+1},}$$
$$\displaystyle{ \vert \,y_{k+1} - p_{k+1} - dA_{k}\varDelta _{k}\vert \leq d\vert b_{k+1}\vert \leq d\beta _{0}. }$$
(2.146)

On the other hand, if dCδ(μ), then it follows from (2.52) that

$$\displaystyle{\vert \,f(\,y_{k}) - p_{k+1} - dA_{k}\varDelta _{k}\vert = \vert \,f(\,p_{k} + d\varDelta _{k}) - f(\,p_{k}) - dA_{k}\varDelta _{k}\vert \leq }$$
$$\displaystyle{ \leq \mu \vert d\varDelta _{k}\vert \leq \mu dC < d }$$
(2.147)

(see (2.145)).

Combining (2.146) and (2.147), we see that

$$\displaystyle{\vert \,y_{k+1} - f(\,y_{k})\vert < d(\beta _{0} + 1),\quad k \in \mathbb{Z},}$$

if d is small enough. Let us emphasize that the required smallness of d depends on β 0, N, and estimates on ∥A k ∥.

Now the assumptions of our lemma imply that there exist sequences x k and t k such that

$$\displaystyle{\vert t_{k} - 1\vert \leq d\beta,\;\vert x_{k} - y_{k}\vert \leq d\beta,\;x_{k+1} =\phi (t_{k},x_{k}),\quad k \in \mathbb{Z}.}$$

If we represent

$$\displaystyle{x_{k} = p_{k} + dc_{k}\mbox{ and }t_{k} = 1 + ds_{k},}$$

then

$$\displaystyle{\vert dc_{k} - d\varDelta _{k}\vert = \vert x_{k} - y_{k}\vert \leq d\beta.}$$

Thus,

$$\displaystyle{ \vert c_{k} -\varDelta _{k}\vert \leq \beta,\quad - N \leq k \leq N. }$$
(2.148)

Clearly,

$$\displaystyle{ \vert s_{k}\vert \leq \beta,\quad k \in \mathbb{Z}. }$$
(2.149)

Define mappings

$$\displaystyle{G_{k}: \mathbb{R} \times \mathbb{R}^{n} \rightarrow \mathbb{R}^{n},\quad k \in \mathbb{Z},}$$

by

$$\displaystyle{G_{k}(t,v) =\phi (1 + t,p_{k} + v) - p_{k+1}.}$$

Then

$$\displaystyle{G_{k}(0,0) = 0,\;D_{t}G_{k}(t,v)\vert _{t=0,v=0} = X(\,p_{k+1}),\;D_{v}G_{k}(t,v)\vert _{t=0,v=0} = A_{k}.}$$

We can write the equality

$$\displaystyle{x_{k+1} =\phi (1 + ds_{k},x_{k})}$$

in the form

$$\displaystyle{\,p_{k+1} + dc_{k+1} =\phi (1 + ds_{k},p_{k} + dc_{k}),}$$

which is equivalent to

$$\displaystyle{ dc_{k+1} = G_{k}(ds_{k},dc_{k}). }$$
(2.150)

Now we fix a sequence of values d = d (m) → 0,  m. Let us denote by c k (m), t k (m), and s k (m) the values c k , t k , and s k defined above and corresponding to d = d (m).

It follows from estimates (2.148) and (2.149) that | c k (m) | ≤ C + β and | s k (m) | ≤ β for all m and − NkN − 1. The second inequality implies that \(\left \vert t_{k}^{(m)}\right \vert \leq 1\) for large m. Hence (passing to a subsequence, if necessary), we can assume that

$$\displaystyle{c_{k}^{(m)} \rightarrow \tilde{ c}_{ k},\;t_{k}^{(m)} \rightarrow \tilde{ t}_{ k},\;s_{k}^{(m)} \rightarrow \tilde{ s}_{ k},\quad m \rightarrow \infty,}$$

for − NkN − 1.

Applying relations (2.150) and (2.149), we can write

$$\displaystyle{d_{m}c_{k+1}^{(m)} = G_{ k}\left (d_{m}s_{k}^{(m)},d_{ m}c_{k}^{(m)}\right ) = A_{ k}d_{m}c_{k+1}^{(m)}+X(\,p_{ k+1})d_{m}s_{k}^{(m)}+o(d_{ m}).}$$

Dividing these equalities by d m , we get the relations

$$\displaystyle{c_{k+1}^{(m)} = A_{ k}c_{k+1}^{(m)} + X(\,p_{ k+1})s_{k}^{(m)} + o(1),\quad - N \leq k \leq N - 1.}$$

Letting m, we arrive at the relations

$$\displaystyle{\tilde{c}_{k+1} = A_{k}\tilde{c}_{k} + X(\,p_{k+1})\tilde{s}_{k},\quad - N \leq k \leq N - 1,}$$

where

$$\displaystyle{\vert \varDelta _{k} -\tilde{ c}_{k}\vert,\vert \tilde{s}_{k}\vert \leq \beta,\quad - N \leq k \leq N - 1,}$$

due to (2.148) and (2.149).

Recall that N was fixed in the above reasoning. Denote the obtained \(\tilde{s}_{k}\) by s k (N). Then \(v_{k}^{(N)} = \varDelta _{k} -\tilde{ c}_{k}\) is a solution of the equations

$$\displaystyle{v_{k+1}^{(N)} = A_{ k}v_{k}^{(N)} + X(\,p_{ k+1})s_{k}^{(N)} + b_{ k+1},\quad - N \leq k \leq N - 1,}$$

such that \(\left \vert v_{k}^{(N)}\right \vert \leq \beta\).

There exist subsequences \(v_{k}^{(\,j_{N})} \rightarrow v_{k}^{{\prime}}\) and \(s_{k}^{(\,j_{N})} \rightarrow s_{k}^{{\prime}}\) as N (we do not assume uniform convergence) such that

$$\displaystyle{v_{k+1}^{{\prime}} = A_{ k}v_{k}^{{\prime}} + X(\,p_{ k+1})s_{k}^{{\prime}} + b_{ k+1},\quad k \in \mathbb{Z},}$$

and | v k |, | s k | ≤ β. The lemma is proved. □

Remark 2.7.1

An analog of Lemma 2.7.2 is valid in the case of a smooth closed manifold M. In this case, we denote \(\mathcal{M}_{k} = T_{p_{k}}M\) and consider the difference equation (2.142) in which \(v_{k},b_{k} \in \mathcal{M}_{k}\).

Proving an analog of Lemma 2.7.2 in the case of a closed manifold (and replacing, for example, the formula y k = p k + k by \(y_{k} =\exp _{p_{k}}(d\varDelta _{k})\), compare with the proof of Lemma 2.3.3 in Sec 2.3), one gets a similar statement with the estimates | s k | ≤ β: = L(2β 0 + 1) and ∥V ≤ 2β (see the original paper [57]).

Thus, in what follows, we refer to Lemma 2.7.2 in the case of a vector field X on a smooth closed manifold M (with \(B =\{ b_{k} \in \mathbb{R}^{n}\}\) replaced by \(B =\{ b_{k} \in \mathcal{M}_{k}\}\) and properly corrected estimates).

2.7.2 Rest Points

In this section, we show that if a vector field has the Lipschitz shadowing property, then its rest points are hyperbolic and isolated in the chain recurrent set. Thus, in what follows we assume that we work with a vector field X on a smooth closed manifold M having the Lipschitz shadowing property.

Lemma 2.7.3

Every rest point of X is hyperbolic.

Proof

Let x 0 be a rest point. Applying an analog of Lemma 2.7.2 for the case of a manifold with p k = x 0 and noting that X( p k ) = 0, we conclude that the difference equation

$$\displaystyle{v_{k+1} = Df(x_{0})v_{k} + b_{k+1}}$$

has a bounded solution for any bounded sequence \(b_{k} \in \mathcal{M}_{k}\) (recall that \(\mathcal{M}_{k} = T_{p_{k}}M\)).

Then it follows from the Maizel’ theorem (see Theorem 2.1.1 of Sect. 2.1) that the constant sequence \(\mathcal{A} =\{ A_{k} = Df(x_{0})\}\) is hyperbolic on \(\mathbb{Z}_{+}\); in particular, every bounded solution of the equation

$$\displaystyle{v_{k+1} = Df(x_{0})v_{k}}$$

tends to 0 as k.

However, if the rest point x 0 is not hyperbolic, then the matrix Df(x 0) has an eigenvalue on the unit circle, in which case the above equation has a nontrivial solution with constant norm. Thus, x 0 is hyperbolic. □

Lemma 2.7.4

Rest points are isolated in the chain recurrent set \(\mathcal{R}(\phi )\) .

Proof

Let us assume that there exists a rest point x 0 that is not isolated in \(\mathcal{R}(\phi )\). First we want to show that there is a homoclinic trajectory x(t) associated with x 0.

Since x 0 is hyperbolic by the previous lemma, there exists a small d > 0 and a number a > 0 such that if \(\mbox{ dist}(\phi (t,y),x_{0}) \leq \mathcal{L}d\) for | t | ≥ a, then ϕ(t, y) → x 0 as | t | → .

Assume that there exists a point \(y \in \mathcal{R}(\phi )\) such that y is arbitrarily close to x 0 and yx 0. Given any ɛ 0, θ > 0 we can find points y 1, , y N and numbers T 0, , T N > θ such that

$$\displaystyle{\mbox{ dist}(\phi (T_{0},y),y_{1}) <\varepsilon _{0},}$$
$$\displaystyle{\mbox{ dist}(\phi (T_{i},y_{i}),y_{i+1}) <\varepsilon _{0},\quad i = 1,\ldots,N,}$$

and

$$\displaystyle{\mbox{ dist}(\phi (T_{N},y_{N}),y_{1}) <\varepsilon _{0}.}$$

Set T = T 0 + + T N and define g on [0, T] by

$$\displaystyle{g^{{\ast}}(t) = \left \{\begin{array}{cl} \phi (t,y), &0 \leq t \leq T_{0}; \\ \phi (t,y_{i}),&T_{0} +\ldots +T_{i-1} < t < T_{0} +\ldots +T_{i}; \\ y, &t = T. \end{array} \right.}$$

Clearly, for any ɛ > 0 we can find ɛ 0 depending only on ɛ and ν (see (2.141)) such that g (t) is an ɛ-pseudotrajectory of ϕ on [0, T].

Now we define

$$\displaystyle{g(t) = \left \{\begin{array}{cl} x_{0}, &t \leq 0; \\ g^{{\ast}}(t),&0 < t \leq T; \\ x_{0}, &t > T. \end{array} \right.}$$

We want to choose y and ɛ in such a way that g(t) is a d-pseudotrajectory of ϕ. We have to show that

$$\displaystyle{ \mbox{ dist}(\phi (t,g(\tau )),g(t+\tau )) < d }$$
(2.151)

for all τ and t ∈ [0, 1].

Clearly, (2.151) holds for (i) τ ≤ −1, (ii) τT, (iii) τ, τ + t ∈ [−1, 0], and (iv) τ, τ + t ∈ [0, T] and ɛ < d.

If τ ∈ [−1, 0] and τ + t > 0, then

$$\displaystyle{\mbox{ dist}(\phi (t,g(\tau )),g(t+\tau )) = \mbox{ dist}(x_{0},g^{{\ast}}(t+\tau )) \leq }$$
$$\displaystyle{\leq \mbox{ dist}(x_{0},\phi (t+\tau,y)) + \mbox{ dist}(\phi (t+\tau,y),g^{{\ast}}(t+\tau )) \leq \nu \mbox{ dist}(x_{ 0},y)+\varepsilon,}$$

where ν is as in (2.141). The last value is less than d if dist(x 0, y) and ɛ are small enough. Note that, for a fixed y, we can decrease ɛ and increase N, T 0, , T N arbitrarily so that g(t) remains a d-pseudotrajectory.

Similarly, (2.151) holds if τ ∈ [0, T] and τ + t > T.

Thus, g(t) is \(\mathcal{L}d\) shadowed by a trajectory x(t) such that \(\mbox{ dist}(x(t),x_{0}) \leq \mathcal{L}d\) if | t | is sufficiently large; hence, x(t) → x 0 as | t | → .

Now we want to show that x(t) is a homoclinic trajectory if d is small enough. For this purpose, we have to show that x(t) ≠ x 0.

There exists an \(\varepsilon _{1} > \mathcal{L}d\) (provided that d is small enough) such that if y does not belong to the local stable manifold of x 0, then dist(ϕ(t 0), y) ≥ ɛ 1 for some t 0 > 0. We can choose T 0 > t 0 (not changing the point y). Then g(t) contains the point g (t 0) = ϕ(t 0, y) whose distance to x 0 is more than \(\mathcal{L}d\). Hence, x(t) contains a point different from x 0, as was claimed.

If y belongs to the local stable manifold of x 0, then it does not belong to the local unstable manifold of x 0. In this case, considering the flow ψ(t, x) = ϕ(−t, x), we can apply the above reasoning to ψ noting that \(\mathcal{R}(\psi ) = \mathcal{R}(\phi )\) and ψ has the Lipschitz shadowing property as well.

Now we show that the existence of this homoclinic trajectory x(t) leads to a contradiction. Set p k = x(k). Since A k X( p k ) = X( p k+1), it is easily verified that if we consider two sequences β k and s k such that

$$\displaystyle{\beta _{k+1} =\beta _{k} + s_{k},\quad k \in \mathbb{Z},}$$

then u k = β k X( p k ) is a solution of

$$\displaystyle{ u_{k+1} = A_{k}u_{k} + X(\,p_{k+1})s_{k},\quad k \in \mathbb{Z}. }$$
(2.152)

In addition, if the sequence s k is bounded, then the sequence β k X( p k ) is bounded as well since X( p k ) → 0 exponentially as | k | → (the trajectory x(t) tends to a hyperbolic rest point as time goes to ±) and the sequence | β k | ∕ | k | is bounded).

By Lemma 2.7.2, for any bounded sequence \(b_{k} \in \mathcal{M}_{k}\) there exists a bounded scalar sequence s k such that Eqs. (2.142) have a bounded solution v k . We have shown that Eqs. (2.152) have a bounded solution u k . Then the sequence w k = v k u k is bounded and satisfies the equations

$$\displaystyle{w_{k} = A_{k}w_{k} + b_{k+1},\quad k \in \mathbb{Z}.}$$

Thus, the sequence \(\mathcal{A} =\{ A_{k}\}\) has the Perron property on \(\mathbb{Z}\). It follows from Theorems 2.1.1 and 2.1.2 that the sequence \(\mathcal{A}\) is hyperbolic both on \(\mathbb{Z}_{+}\) and \(\mathbb{Z}_{-}\) and the corresponding spaces S 0 + and U 0 are transverse. But this leads to a contradiction since

$$\displaystyle{\mbox{ dim}S_{0}^{+} + \mbox{ dim}U_{ 0}^{-} = \mbox{ dim}M}$$

(because dimS 0 + equals the dimension of the stable manifold of the hyperbolic rest point x 0 and dimU 0 equals the dimension of its unstable manifold), while any of the spaces S 0 + and U 0 contains the nonzero vector X( p 0). The lemma is proved. □

2.7.3 Hyperbolicity of the Chain Recurrent Set

We have shown that rest points of ϕ are hyperbolic and isolated in the chain recurrent set \(\mathcal{R}(\phi )\). Since M is compact, this implies that the set \(\mathcal{R}(\phi )\) is the union of a finite set of hyperbolic rest points and a compact set (let us denote it Σ) on which the vector field X is nonzero.

To show that \(\mathcal{R}(\phi )\) is hyperbolic, it remains to show that the set Σ is hyperbolic.

Consider the subbundle \(\mathcal{V }(\varSigma )\) of the tangent bundle TM | Σ defined in Sect. 1.3 before Theorem 1.3.15.

Let x(t) be a trajectory in Σ. Let us introduce the following notation. Put p k = x(k) and let P k = P( p k ) and V k = V ( p k ) (recall that P(x) is the orthogonal projection in T x M with kernel spanned by X(x) and V (x) is the orthogonal complement to X(x) in T x M). Introduce the operators

$$\displaystyle{B_{k} = P_{k+1}A_{k}:\; V _{k} \rightarrow V _{k+1}}$$

(recall that A k = Df( p k )).

Lemma 2.7.5

For every bounded sequence b k V k there exists a bounded solution v k V k of

$$\displaystyle{ v_{k+1} = B_{k}v_{k} + b_{k+1},\quad k \in \mathbb{Z}. }$$
(2.153)

Proof

Fix a bounded sequence b k V k . There exist bounded sequences s k of scalars and w k of vectors in \(T_{p_{k}}M\) such that

$$\displaystyle{ w_{k+1} = A_{k}w_{k} + X(\,p_{k+1})s_{k} + b_{k+1},\quad k \in \mathbb{Z}, }$$
(2.154)

(see the remark after Lemma 2.7.2).

Note that A k X( p k ) = X( p k+1). Since (Id − P k )v ∈ {X( p k )} for \(v \in \mathcal{M}_{k}\), we see that

$$\displaystyle{\,P_{k+1}A_{k}(\mbox{ Id} - P_{k}) = 0,}$$

which gives us the equality

$$\displaystyle{ P_{k+1}A_{k} = P_{k+1}A_{k}P_{k}. }$$
(2.155)

The properties of the set Σ imply that the projections P k are uniformly bounded.

Multiplying (2.154) by P k+1, taking into account the equalities P k+1 X( p k+1) = 0 and P k+1 b k+1 = b k+1, and applying (2.155), we conclude that v k = P k w k is the required bounded solution of (2.153). The lemma is proved. □

It follows from the above lemma that if we fix a trajectory x(t) in Σ and consider the corresponding sequence of operators \(\mathcal{B} =\{ B_{k}\}\), then \(\mathcal{B}\) has the Perron property.

By Theorems 2.1.1 and 2.1.2, the sequence \(\mathcal{B}\) is hyperbolic both on \(\mathbb{Z}_{-}\) and \(\mathbb{Z}_{+}\) and the corresponding spaces \(U_{0}^{-}(\mathcal{B})\) and \(S_{0}^{+}(\mathcal{B})\) are transverse.

Consider the mapping π on the normal bundle \(\mathcal{V }(\varSigma )\) defined in Sect. 1.3. Recall that

$$\displaystyle{\pi (x,v) = (\,f(x),B(x)v),\mbox{ where }B(x) = P(\,f(x))Df(x)}$$

(see Sect. 1.3).

In fact, we have shown that π satisfies an analog of the strong transversality condition.

The same reasoning as in the proof of Lemma 2.2.5 shows that the dual mapping π does not have nontrivial bounded trajectories. It is easy to show that if the flow ϕ has the shadowing property, then its nonwandering set coincides with its chain recurrent set.

Hence, we can repeat the reasoning of the proof of Theorem 2.2.2 to conclude that the mapping π is hyperbolic.

It remains to refer to Theorem 1.3.15 to conclude that Σ is a hyperbolic set of the flow ϕ.

2.7.4 Transversality of Stable and Unstable Manifolds

Let x(t) be a trajectory that belongs to the intersection of the stable and unstable manifolds of two trajectories, x +(t) and x (t), respectively, lying in the chain recurrent set of ϕ.

Without loss of generality, we may assume that

$$\displaystyle{\mbox{ dist}(x(0),x_{+}(0)) \rightarrow 0,\quad t \rightarrow \infty,}$$

and

$$\displaystyle{\mbox{ dist}(x(0),x_{-}(0)) \rightarrow 0,\quad t \rightarrow -\infty.}$$

Denote \(p_{k} = x(k),\;k \in \mathbb{Z}\); let W s( p 0) and W u( p 0) be the stable and unstable manifolds of p 0, respectively. Then, of course, W s( p 0) = W s(x +(0)) and W u( p 0) = W u(x (0)). Denote by E s and E u the tangent spaces of W s( p 0) and W u( p 0) at p 0.

We use the notation introduced before Lemma 2.7.5.

By Lemma 2.7.5, for any bounded sequence b k V k there exists a bounded solution v k V k of (2.153). By the Maizel’ theorem (Theorem 2.1.1), the sequence B k is hyperbolic on \(\mathbb{Z}_{-}\) and \(\mathbb{Z}_{+}\).

By the Pliss theorem (Theorem 2.1.2),

$$\displaystyle{ \mathcal{E}^{s} + \mathcal{E}^{u} = V _{ 0}, }$$
(2.156)

where

$$\displaystyle{\mathcal{E}^{s} =\{ w_{ 0}:\; w_{k+1} = B_{k}w_{k},\;\vert w_{k}\vert \rightarrow 0,\;k \rightarrow \infty \}}$$

and

$$\displaystyle{\mathcal{E}^{u} =\{ w_{ 0}:\; w_{k+1} = B_{k}w_{k},\;\vert w_{k}\vert \rightarrow 0,k \rightarrow -\infty \}.}$$

Clearly, it follows from the hyperbolicity of the sequence B k on \(\mathbb{Z}_{-}\) and \(\mathbb{Z}_{+}\) that the following equalities hold:

$$\displaystyle{\mathcal{E}^{s} =\{ w_{ 0}:\; w_{k+1} = B_{k}w_{k},\;\sup _{k\geq 0}\vert w_{k}\vert < \infty \}}$$

and

$$\displaystyle{\mathcal{E}^{u} =\{ w_{ 0}:\; w_{k+1} = B_{k}w_{k},\;\sup _{k\leq 0}\vert w_{k}\vert < \infty \}.}$$

We claim that

$$\displaystyle{ \mathcal{E}^{s} \subset E^{s}\mbox{ and }E^{u} \subset \mathcal{E}^{u}. }$$
(2.157)

First we note that (2.157) implies the desired transversality of W s( p 0) and W u( p 0) at p 0.

Indeed, combining equality (2.156) with inclusions (2.157) and the trivial relations

$$\displaystyle{E^{s} = V _{ 0} \cap E^{s} +\{ X(\,p_{ 0})\}\mbox{ and }E^{u} = V _{ 0} \cap E^{u} +\{ X(\,p_{ 0})\},}$$

we conclude that

$$\displaystyle{E^{s} + E^{u} = T_{ p_{0}}M,}$$

which gives us the transversality of W s( p 0) and W u( p 0) at p 0.

Thus, it remains to prove inclusions (2.157). We prove the first inclusion; for the second one, the proof is similar.

  • Case 1: The limit trajectory x 0(t) = x 0 is a rest point of X. In this case, the stable manifold of the rest point x 0 in the flow ϕ coincides with the stable manifold of the fixed point x 0 for the time-one diffeomorphism f(x) = ϕ(1, x).

It is clear that if p k is a trajectory of f belonging to the stable manifold of x 0, then the tangent space to the stable manifold at p 0 is the subspace E s of the initial values of bounded solutions of

$$\displaystyle{ v_{k+1} = A_{k}v_{k},\quad k \geq 0. }$$
(2.158)

Let us prove that \(\mathcal{E}^{s} \subset E^{s}\). Fix an arbitrary sequence w k such that w k+1 = B k w k and \(w_{0} \in \mathcal{E}^{s}\). Consider the sequence

$$\displaystyle{v_{k} = \lambda _{k}X(\,p_{k})/\vert X(\,p_{k})\vert + w_{k},}$$

where the λ k satisfy the relations

$$\displaystyle{ \lambda _{k+1} = \frac{\vert X(\,p_{k+1})\vert } {\vert X(\,p_{k})\vert } \lambda _{k} + \frac{X(\,p_{k+1})^{{\ast}}} {\vert X(\,p_{k+1})\vert }A_{k}w_{k} }$$
(2.159)

(we denote by X the row-vector corresponding to the column-vector X) and λ 0 = 0. It is easy to see that the sequence v k satisfies (2.158).

Since x(t) is in the stable manifold of the hyperbolic rest point x 0, there exist positive constants K and α such that

$$\displaystyle{\left \vert \frac{dx} {dt} (t)\right \vert \leq K\exp (\alpha (t - s))\left \vert \frac{dx} {dt} (s)\right \vert,\quad 0 \leq s \leq t.}$$

This implies that

$$\displaystyle{\left \vert X(\,p_{k})\right \vert \leq K\exp (\alpha (k - m))\left \vert X(\,p_{m})\right \vert,\quad 0 \leq m \leq k.}$$

Thus, the scalar difference equation

$$\displaystyle{\lambda _{k+1} = \frac{\vert X(\,p_{k+1})\vert } {\vert X(\,p_{k})\vert } \lambda _{k}}$$

is hyperbolic on \(\mathbb{Z}_{+}\) and is, in fact, stable. Since the second term on the right in (2.159) is bounded as k (recall that we take \(w_{0} \in \mathcal{E}^{s}\)), it follows that the λ k are bounded for any choice of λ 0.

We conclude that v k is a bounded solution of (2.158), and v 0 = w 0E s. Thus, we have shown that \(\mathcal{E}^{s} \subset E^{s}\), which completes the proof in Case 1.

  • Case 2: The limit trajectory is in the set Σ (the chain recurrent set minus rest points). We know that the set Σ is hyperbolic. Our goal is to find the intersection of its stable manifold near p 0 = x(0) with the cross-section at p 0 orthogonal to the vector field (in local coordinates generated by the exponential mapping). To do this, we discretize the problem and note that there exists a number σ > 0 such that a point p close to p 0 belongs to W s( p 0) if and only if the distances of the consecutive points of intersections of the positive semitrajectory of p to the points p k do not exceed σ.

For suitably small μ > 0 we find all the sequences of numbers t k and vectors z k V k (recall that V k is the orthogonal complement to {X( p k )} at p k ) such that

$$\displaystyle{\vert t_{k} - 1\vert \leq \mu,\;\vert z_{k}\vert \leq \mu,\;y_{k+1} =\phi (t_{k},y_{k}),\quad k \geq 0,}$$

where y k = p k + z k .

Thus, we have to solve the equations

$$\displaystyle{\,p_{k+1} =\phi (t_{k},p_{k} + v_{k}),\quad k \geq 0,}$$

for numbers t k and vectors z k V k such that | t k − 1 | ≤ μ and | z k | ≤ μ.

We reduce this problem to an equation in a Banach space. It was mentioned above that the sequence {B k } generating the difference equation

$$\displaystyle{z_{k} = B_{k}z_{k},\quad k \geq 0,}$$

(where B k = P k+1 A k and P k is the orthogonal projection with range V k ) is hyperbolic on \(\mathbb{Z}_{+}\). Denote by Q k :   V k V k the corresponding projections to the stable subspaces and by \(\mathcal{R}(Q_{0})\) the range of Q 0 (note that \(\mathcal{R}(Q_{0}) = \mathcal{E}^{s}\)).

Fix a positive number μ 0 and denote by \(\mathcal{V }\) the space of sequences

$$\displaystyle{\mathcal{V } =\{ z_{k} \in V _{k}:\; \vert z_{k}\vert \leq \mu _{0},\;k \in \mathbb{Z}_{+}\}.}$$

Let \(l^{\infty }\left (\mathbb{Z}_{+},\{\mathcal{M}_{k+1}\}\right )\) be the space of sequences \(\{\zeta _{k} \in \mathcal{M}_{k+1}:\; k \in \mathbb{Z}_{+}\}\) with the usual norm.

Define a C 1 function

$$\displaystyle{G:\; [1 -\mu _{0},1 +\mu _{0}]^{\mathbb{Z}_{+} } \times \mathcal{V } \times \mathcal{R}(Q_{0}) \rightarrow l^{\infty }(\mathbb{Z}_{ +},\{\mathcal{M}_{k+1}\}) \times \mathcal{R}(Q_{0})}$$

by

$$\displaystyle{G(t,z,\eta ) = (\{\,p_{k+1} + z_{k+1} -\phi (t_{k},p_{k} + z_{k})\},Q_{0}z_{0}-\eta ).}$$

This function is defined if μ 0 is small enough.

We want to solve the equation

$$\displaystyle{G(t,z,\eta ) = 0}$$

for (t, z) as a function of η. It is clear that

$$\displaystyle{G(1,0,0) = 0,}$$

where the first argument of G is {1, 1, }, the second argument is {0, 0, }, and the right-hand side is ({0, 0, }, 0).

To apply the implicit function theorem, we must verify that the operator

$$\displaystyle{T = \frac{\partial G} {\partial (t,z)}(1,0,0)}$$

is invertible.

First we note that if \((s,w) \in l^{\infty }(\mathbb{Z}_{+},\{\mathcal{M}_{k+1}\}) \times \mathcal{V }\), then

$$\displaystyle{T(s,w) = (\{w_{k+1} - X(\,p_{k+1})s_{k} - A_{k}w_{k}\},Q_{0}w_{0}).}$$

To show that T is invertible, we have to show that the equation

$$\displaystyle{T(s,w) = (g,\eta )}$$

has a unique solution for any \((g,\eta ) \in l^{\infty }(\mathbb{Z}_{+},\{\mathcal{M}_{k+1}\}) \times \mathcal{R}(Q_{0})\). Thus, we have to solve the equation

$$\displaystyle{ w_{k+1} = A_{k}w_{k} + X(\,p_{k+1})s_{k} = g_{k},\quad k \geq 0, }$$
(2.160)

subject to the condition

$$\displaystyle{Q_{0}w_{0} =\eta.}$$

If we multiply Eq. (2.160) by X( p k+1) and solve for s k , we get the equalities

$$\displaystyle{s_{k} = - \frac{X(\,p_{k+1})^{{\ast}}} {\vert X(\,p_{k+1})\vert ^{2}}[A_{k}w_{k} + g_{k}],\quad k \geq 0,}$$

and if we multiply Eq. (2.160) by P k+1, we get the equalities

$$\displaystyle{w_{k+1} = P_{k+1}A_{k}w_{k} + P_{k+1}g_{k} = B_{k}w_{k} + P_{k+1}g_{k},\quad k \geq 0.}$$

Now we know that the last equations have a unique bounded solution w k V k ,  k ≥ 0, that satisfies Q 0 w 0 = η. Thus, T is invertible.

Hence, we can apply the implicit function theorem to show that there exists a μ > 0 such that if | η | is sufficiently small, then the equation G(t, z, η) = 0 has a unique solution (t(η), z(η)) such that ∥t − 1∥ μ and ∥z μ. Moreover, t(0) = 1, z(0) = 0, and the functions t(η) and z(η) are of class C 1.

The points p 0 + z 0(η) form a submanifold containing p 0 and contained in W s( p 0). Thus, the range of the derivative z 0 (0) is contained in E s.

Take an arbitrary vector \(\xi \in \mathcal{E}^{s}\) and consider \(\eta =\tau \xi,\;\tau \in \mathbb{R}\). Differentiating the equalities

$$\displaystyle{\,p_{k+1} + z_{k+1}(\tau \xi ) =\phi (t_{k}(\tau \xi ),p_{k} + z_{k}(\tau \xi )),\quad k \geq 0,}$$

and

$$\displaystyle{Q_{0}(\tau \xi ) =\tau \xi }$$

with respect to τ at τ = 0, we see that

$$\displaystyle{s_{k} = \frac{\partial t_{k}} {\partial \eta } \vert _{\eta =0}\xi \mbox{ and }w_{k} = \frac{\partial z_{k}} {\partial \eta } \vert _{\eta =0}\xi \in V _{k}}$$

are bounded sequences satisfying the equalities

$$\displaystyle{w_{k+1} = A_{k}w_{k} + X(\,p_{k+1})s_{k}\mbox{ and }Q_{0}w_{0} =\xi.}$$

Multiplying by P k+1, we conclude that

$$\displaystyle{w_{k+1} = B_{k}w_{k}\mbox{ and }Q_{0}w_{0} =\xi.}$$

It follows that \(w_{0} \in \mathcal{E}^{s} = \mathcal{R}(Q_{0})\). Then w 0 = Q 0 w 0 = ξ.

We have shown that the range of z 0 (0) is exactly \(\mathcal{E}^{s}\). Thus, \(\mathcal{E}^{s} \subset E^{s}\).

Historical Remarks

Theorem 2.7.1 was published by K. Palmer, the first author, and S. B. Tikhomirov in [57].