2 Preliminaries

The dynamical system $(X,{\mathcal {B}},\mu , T)$ is said to be non-singular if the map $T:X\to X$ is a non-singular transformation of $(X,\mu )$ , that is, for any $N\in {\mathcal {B}}$ , $\mu (T N)=0$ if and only if $\mu (N)=0$ . Recall that the system is measure-preserving if $\mu (A)=\mu (T A)$ for all $A\in {\mathcal {B}}$ . By the Poincaré recurrence lemma, measure-preserving transformations are conservative.

The structure of non-singular transformations is given by the Hopf decomposition theorem, a proof of which can be found in [Reference Aaronson1].

The set C is called the conservative part of T. If the non-singular transformation is invertible, so that T and its inverse $T^{-1}$ are measurable, then we have both $\mu \circ T^{-1}\sim \mu $ and $\mu \circ T\sim \mu $ . However, we do not assume further the transformation is invertible.

We will denote the Radon–Nikodým derivative ${d(\mu \circ T^{i})}/{d{\kern-1pt}\mu }$ by $\omega _{i}$ . Note that the Radon–Nikodým derivatives must satisfy the cocycle identity

$$ \begin{align*}\omega_{i+j}(x)=\omega_{i}(x)\omega_{j}(T^{i}x)\end{align*} $$

for almost every x and for every $i,j \in {\mathbb {Z}}$ . Clearly, T is measure-preserving if and only if $\omega _{i}(x)=1$ for almost every x for all i.

It follows that for every $f\in L^{1}(X,\mu )$

$$ \begin{align*} \int_{X}f(x)\,d{\kern-1pt}\mu(x)=\int_{X}f(Tx)\omega_{1}(x)\,d{\kern-1pt}\mu(x)=\int_{X}f(T^{n}x)\omega_{n}(x)\,d{\kern-1pt}\mu(x). \end{align*} $$

If $f_{n}=\sum _{i=0}^{n-1}f(T^{i}x)\omega _{i}(x),n\geq 1$ , where $\omega _{0}(x)=1$ , it is easy to show that $f_{m+n}(x)=f_{n}(x)+\omega _{n}(x)f_{m}(T^{n}x)$ . The Hurewicz ergodic theorem [Reference Hurewicz7] is a generalization of the Birkhoff ergodic theorem to the setting of non-singular conservative transformations.

Theorem 3. (Hurewicz ergodic theorem)

Let $(X,{\mathcal {B}},\mu )$ be a probability space, and $T:X\to X$ a non-singular conservative transformation. If $f\in L^{1}(\mu )$ , then

$$ \begin{align*} \lim_{n\to\infty} \frac{\sum_{i=0}^{n-1}f(T^{i}x)\omega_{i}(x)}{\sum_{i=0}^{n-1}\omega_{i}(x)}=f_{\ast}(x) \end{align*} $$

exists $\mu $ -a.e. Furthermore, $f_{\ast }(x)$ is T- invariant and

$$ \begin{align*} \int_{X}f(x)\,d{\kern-1pt}\mu(x)=\int_{X}f_{\ast}(x)\,d{\kern-1pt}\mu(x). \end{align*} $$

Note that if T is measure-preserving, the left-hand side becomes

$$ \begin{align*} \lim _{n\to \infty } ({1}/{n})\sum _{i=0}^{n-1} f(T^{i}x), \end{align*} $$

and so we recover the Birkhoff theorem.

3 Non-singular Kingman theorem

Thus, let T be a conservative non-singular transformation of a measure space $(X, {\mathcal {B}}, \mu )$ , and denote by $\omega _i$ the Radon–Nikodým derivative $({d{\kern-1pt}\mu \circ T^{i}})/{ d{\kern-1pt}\mu }$ .

Definition 1. We say that $\{f_n \}$ in $L^1(X, \mu )$ is a subadditive sequence for T if, for all integers m and n,

$$ \begin{align*} f_{m+n}(x)\leq f_{n}(x)+\omega_{n}(x)f_{m}(T^{n}x). \end{align*} $$

It is easy to see that if f is integrable, then

$$ \begin{align*} f_{n}(x) = \sum_{i=0}^{n-1}f(T^{i}x)\omega_{i}(x). \end{align*} $$

is subadditive.

Similarly, we say that $\{f_n \}$ in $L^1(X, \mu )$ is superadditive for T if, for all integers m and n,

$$ \begin{align*} f_{m+n}(x)\geq f_{n}(x)+\omega_{n}(x)f_{m}(T^{n}x). \end{align*} $$

Observe that $f_{n}$ is a superadditive sequence if and only if $-f_{n}$ is a subadditive sequence.

We now state the non-singular Kingman theorem,

Theorem 4. (Non-singular Kingman ergodic theorem)

Let $(X,{\mathcal {B}},\mu )$ be a probability space, and $T:X\to X$ a non-singular conservative transformation. Let $f_{n}\in L^{1}$ be a sequence of functions satisfying the subadditivity relation $f_{m+n}(x)\leq f_{n}(x)+\omega _{n}(x)f_{m}(T^{n}x)$ for almost all $x\in X$ . Then

$$ \begin{align*} \lim_{n\to\infty} \frac{f_{n}(x)}{\sum_{i=0}^{n-1}\omega_{i}(x)}=f_{\ast}(x)< \infty \end{align*} $$

exists $\mu $ -a.e. Furthermore, $f_{\ast }(x)$ is T-invariant and

$$ \begin{align*} \int_{X}f(x)\,d{\kern-1pt}\mu(x)=\int_{X}f_{\ast}(x)\,d{\kern-1pt}\mu(x). \end{align*} $$

This theorem follows easily from the following fact, which follows from the maximal ergodic theorem:

$$ \begin{align*} \int_{X}\limsup_{n\to\infty}\frac{f_{n}}{\sum_{i=0}^{n-1}\omega_{i}(x)}\,d{\kern-1pt}\mu\leq \int L \,d{\kern-1pt}\mu \leq \int_{X}\liminf_{n\to\infty} \frac{f_{n}}{\sum_{i=0}^{n-1}\omega_{i}(x)}\,d{\kern-1pt}\mu \end{align*} $$

where L is

$$ \begin{align*}\lim_{n\to\infty}\frac{f_{n}}{\sum_{i=0}^{n-1}\omega_{i}(x)}.\end{align*} $$

Proof. Theorem 3.4 of [Reference Silva and Thieullen14] states that if $\{f_n\}$ is subadditive and $\{g_n\}$ is superadditive, the limit

$$ \begin{align*} \lim_{n\to \infty} \frac{f_n}{g_n} = \frac{\lim_{n \to \infty}(1/n) \mathbb{E}_{h\mu} [ f_n / h |\mathcal{I}]}{\lim_{n \to \infty}(1/n) \mathbb{E}_{h\mu} [ g_n / h |\mathcal{I}]} \end{align*} $$

for any positive $\mu $ -integrable function h (where $\mathcal {I}$ is the invariant $\sigma $ -algebra). In our case, we may take $h=1$ , since X is a probability space, and noting that $g_n(x) = \sum _{i=0}^{n-1}\omega _{i}(x)$ is a superadditive sequence of non-negative functions, the result follows.

We would like to thank the referee for pointing out this elegant proof of the theorem. Silva and Thieullen’s proof of their Theorem 3.4 uses a maximal function argument: their Lemma 2.4 extends the maximal ergodic theorem to subadditive sequences. Using it, we obtain the following maximal function estimate for non-singular actions.

Theorem 5. Suppose that $(f_{n})$ is a subadditive sequence of functions satisfying the integrability condition $f_{1}^{+}\in L^{1}$ , and that $p\geq 1$ an integer. Define

$$ \begin{align*} A = \{ x\in X: f_{k}(x)\geq 0 \text{ for all }1\leq k\leq p \} \end{align*} $$

and

$$ \begin{align*} B = \{ x\in X: f_{k}(x)> 0 \text{ for some }1\leq k\leq p \}. \end{align*} $$

Suppose further that for all integers $n\geq p$ and for almost every $x\in X$ we have

$$ \begin{align*}f_{n}\leq \sum^{n-1}_{i=0}f_{1}\circ T^{i}\chi_{A}\circ T^{i}\omega_{i}+\sum^{n-1}_{i=n-p}\|f_{1}\circ T^{i}\|\omega_{i}\end{align*} $$

and

$$ \begin{align*}f_{n}^{+}\leq \sum^{n-1}_{i=0}f_{1}\circ T^{i}\chi_{B}\circ T^{i}\omega_{i}+\sum^{n-1}_{i=n-p}\|f_{1}\circ T^{i}\|\omega_{i}.\end{align*} $$

Then

$$ \begin{align*}\inf_{n\geq 1}\bigg(\frac{1}{n}\bigg)\int f_{n} \,d{\kern-1pt}\mu=\lim_{n\to\infty}\bigg(\frac{1}{n}\bigg)\int f_{n}\,d{\kern-1pt}\mu\leq\int_{A}f_{1}\,d{\kern-1pt}\mu,\end{align*} $$

$$ \begin{align*}0\leq \lim_{n\to\infty}\bigg(\frac{1}{n}\bigg)\int f_{n}^{+}\,d{\kern-1pt}\mu\leq\int_{B}f_{1}\,d{\kern-1pt}\mu .\end{align*} $$

Many proofs of the Kingman theorem deduce it from a maximal inequality; see [Reference Jones8, Reference Katznelson and Weiss9]. There are other types of proofs [Reference Avila and Bochi3, Reference Steel15] which do not rely on a maximal inequality. One can also follow these approaches, replacing the quantity $1/n$ with ${1}/{\Omega _n(x)}$ , where $\Omega _n(x)=\sum _{i=0}^{n-1}\omega _{i}(x)$ , to find an alternative proof of Theorem 4.

Note that in the measure-preserving case, we have $\Omega _n(x) = n$ , which gives us back the standard Kingman theorem, Theorem 1: our theorem shows how to replace the quantity $1/n$ with ${1}/{\Omega _n(x)}$ , which is the key to proving the multiplicative ergodic theorem below.

In the case where the measure $\mu $ is non-singular and has critical dimension $\alpha \in [0,1]$ (see [Reference Dooley and Mortiss5]), we have that ${\Omega _n(x)}/{n^\alpha }$ is non-zero a.e., and the conclusion of Theorem 6 is equivalent to $\lim _{n\to \infty } ({f_{n}(x)}/{n^\alpha })=f(x)< \infty $ .

4 The multiplicative ergodic theorem

We now introduce the notion of cocycles with values in $GL(d)$ of a non-singular transformation T of $(X, \mathcal {B}, \mu )$ ; see [Reference Schmidt13]. A cocycle with respect to the action of T is a function $\Phi : {\mathbb {N}} \times X \to GL(d)$ satisfying $\Phi (n+m,x)=\Phi (n,x)\Phi (m,T^{n-1}x) $ .

Cocycles can be generated by choosing a (random) $d\times d$ matrix, $A(x)$ , for each $x\in X$ , and defining

$$ \begin{align*}\Phi(n,x)=A(x)\times A(T x)\times A(T^{2}x) \cdots \times A(T^{n-1}x).\end{align*} $$

It is easy to see that this formula defines a cocycle. We will say that $A(x)$ is the generator of $\Phi $ .

The operator norm of a square matrix A of dimension d is defined as

$$ \begin{align*}\|A\|=\sup\bigg \{\frac{\|Av\|}{\|v\|}:v\in {\mathbb{R}}^{d}\backslash\{0\} \bigg\}.\end{align*} $$

It follows directly from the definition that the norm of the product of two matrices is less than or equal to the product of the norms of those matrices. Thus

$$ \begin{align*}\|\Phi(n,x)\|\leq \|A(x)\|\|A(T x)\| \cdots \|A(T^{n-1}x)\|.\end{align*} $$

If T is measure-preserving, the Furstenberg–Kesten theorem is an application of the Kingman subadditive ergodic theorem, applied to the subadditive sequence:

$$ \begin{align*}\log \|\Phi(n+m,x)\|\leq \log\|\Phi(n,x)\|+\log\|\Phi(m,T^{m}x)\|.\end{align*} $$

In the non-singular case, we define a (non-singular) subadditive sequence by

$$ \begin{align*}\log\|\Phi(n,x)\|\leq \sum_{i=0}^{n-1} \omega_{i} \log \|A(T^{i}x)\|.\end{align*} $$

We define singular values and exterior powers before we introduce the theorem.

Theorem 6. (Non-singular Furstenberg–Kesten theorem)

Let $\Phi $ be a linear cocycle with one side in discrete time over the non-singular dynamical system $(\Omega ,\mathcal {F},\mu ,T)$ . Assume that the generator $A:X \to Gl(d,{\mathbb {R}})$ of $\Phi $ satisfies

$$ \begin{align*}\log^{+}\|A\| \in L^{1}\end{align*} $$

Then the following statements hold. $(1)$ For each $k=1,\ldots ,d$ , the sequence

$$ \begin{align*}f_{n}^{k}(x)=\log \lVert\land^{k}\Phi(n,x)\rVert,\,n\in{\mathbb{N}},\end{align*} $$

is subadditive and $f_{1}^{k+}\in L^{1}(X,\mathcal {F},\mu )$ . That is,

$$ \begin{align*}f_{n+m}^{k}(x)\leq f_{m}^{k}(x)+ f_{n}^{k}(T^{m}x)\omega_{m}.\end{align*} $$

$(2)$ There is an invariant set $\bar {\Omega }$ of full measure and measurable functions $\gamma ^{k}:X \to {\mathbb {R}}$ with $\gamma ^{k+}\in L^{1}(X,\mathcal {F},\mu )$ :

$$ \begin{align*}\gamma^{k}(x)=\lim_{n\to\infty} \frac{\log\lVert\land^{k}\Phi(n,x)\rVert}{\sum_{i=0}^{n-1}\omega_{i}(x)}\end{align*} $$

and

$$ \begin{align*}\gamma^{k}(T x)=\gamma^{k}(x),\quad\gamma^{k+l}(x)\leq \gamma^{k}(x)+\gamma^{l}(x).\end{align*} $$

Let $\Lambda _{k}$ be the function defined by $\Lambda _{k}=\gamma ^{k+1}-\gamma ^{k},$ and let $\delta _k$ be the corresponding singular value of $\Phi (n,x)$ . Then

$$ \begin{align*}\Lambda_{k}=\lim_{n\to\infty}\frac{\log \delta_{k}(\Phi(n,x))}{\sum_{i=0}^{n-1}\omega_{i}(x)}.\end{align*} $$

Proof. Note that $A(x)=\Phi (1,x)$ . For all k,

$$ \begin{align*}f_{n}^{k}(x)=\log\lVert\land^{k}\Phi(n,x)\rVert\end{align*} $$

is a subadditive sequence, and

$$ \begin{align*}f_{n+1}^{k}(x)=\log\lVert\land^{k}\Phi(n+1,x)\rVert\leq f_{n}^{k}(T x)\omega_{1}(x)+\log\lVert\land^{k}A(x)\rVert .\end{align*} $$

Hence subadditivity of $ f_{n}^{k}(x)$ follows. By Theorem 4, we have

$$ \begin{align*}\gamma^{k}(x)=\lim_{n\to\infty} \frac{\log\lVert\land^{k}\Phi(n,x)\rVert}{\sum_{i=0}^{n-1}\omega_{i}(x)}.\end{align*} $$

Since $\lVert \land ^{k+l}\Phi (n,x)\rVert \leq \lVert \land ^{k}\Phi (n,x)\rVert \lVert \land ^{l}\Phi (n,x)\rVert $ , $\gamma ^{k}$ is a subadditive sequence. For $k=1,\ldots ,d$ ,

$$ \begin{align*}\log\lVert\land^{k}\Phi(n,x)\rVert=\sum_{i=1}^{k}\log\delta_{i}(\Phi (n,x))\end{align*} $$

where $\delta _{i}$ is the corresponding singular value of $\Phi (n,x)$ .

We now consider the behaviour of $ \|\Phi (n,x)v\|$ for $v \in {\mathbb {R}}^{d}$ as $n\to \infty $ . If $A \in M_d({\mathbb {R}})$ with transpose $A^*$ , both $A^{*}A$ and $A A^{*}$ are symmetric and positive semidefinite. Any positive semidefinite and symmetric matrix S may be written in the form

$$ \begin{align*}S=C^{-1}DC\end{align*} $$

where D is diagonal with non-negative entries in non-deceasing order and C is orthogonal.

The polar decomposition of a matrix A is

$$ \begin{align*}A=C(AA^{*})^{{1}/{2}}C^{\prime}=C^{\prime\prime}(A^{*}A)^{{1}/{2}}\end{align*} $$

where $C^{\prime },C^{\prime \prime }$ are orthogonal matrices. Applying the polar decomposition to $\Phi (n,x)$ in the theorem, we obtain

$$ \begin{align*}\Phi(n,x) \approx C^{\prime\prime}_{n} A^{n}(x)\end{align*} $$

for some orthogonal matrix $C^{\prime \prime }_{n}$ . Since orthogonal matrices are isometries, we have $\|C^{\prime \prime }_{n} v\|=\|v\|$ . Thus

$$ \begin{align*}\| \Phi(n,x)v\|=\|A^{n}(x)v\|.\end{align*} $$

Returning to the symmetric matrix $\Phi (n,x)^{*}\Phi (n,x)$ , we know that $\Phi (n,x)^{*}\Phi (n,x)=C_{n}^{*}D_{n}C_{n}$ , and $\Phi (n,x)=L_{n}(D_{n})^{1/2}C_{n}$ , and hence

$$ \begin{align*}\lim_{n\to \infty}(\Phi(n,x)^{*}\Phi(n,x))^{{1}/{2n}}=\lim_{n\to \infty}C_{n}^{*}(D_{n})^{{1}/{2n}}C_{n}.\end{align*} $$

If the limit exists, then there are an orthogonal matrix $C=\lim C_{n}$ and a diagonal matrix $D=\lim D_{n}^{{1}/{2n}}$ . By Theorem 5, we see that $\lim _{n\to \infty } ({\log \lVert \land ^{k}\Phi (n,x)\rVert }/{\sum _{i=0}^{n-1}\omega _{i}(x)})$ converges for all k to $-\infty $ or a finite limit. Hence $\lim _{n\to \infty }({ \log \delta _{i} (\Phi (n,x))}/{\sum _{i=0}^{n-1}\omega _{i}(x)})$ converges to a finite limit for every i. Now we can assume that $D_{n}^{{1}/{2n}} $ converges as $n \to \infty $ .

By the monotonicity of $\Lambda ^{i}$ ,

$$ \begin{align*}\Lambda^{r}(x) \geq \cdots \geq\Lambda^{1}(x).\end{align*} $$

There is a unique partition I, given by

$$ \begin{align*} I=\{1=i_{2}<i_{1}<\cdots<i_{p}<i_{p+1}=r+1\}, \end{align*} $$

such that $ \Lambda ^{i_{q}}= \Lambda ^{i_{q+1}-1} < \Lambda ^{i_{q+1}}$ . This partition splits $\{1,2,\ldots ,r\}$ into finitely many intervals $[i_{q},i_{q+1}-1]$ . If $ \Lambda ^{i}= \Lambda ^{j}$ , then they belong to same interval. Let $\Sigma (I,q)(x)$ be a vector subspace of ${\mathbb {R}}^{r}$ ; it is a union of the zero vector 0 and the set of all eigenvectors corresponding to eigenvalues smaller than or equal to $\Lambda ^{i_{q+1}-1}$ . It is easily seen that $\Sigma (I,0)(x)$ is $\{0\}$ and $\Sigma (I,p)(x)$ is ${\mathbb {R}}^{r}$ . We see that $C_{n}^{-1}e_{i}$ is an eigenvector of $(\Phi (n,x)^{*}\Phi (n,x))^{1/2}$ with eigenvalue $\delta _{i}^{2}$ . We know that $D_{n}^{{1}/{2n}}$ converges, but the question is whether the vector space spanned by $C_{n}^{-1}e_{i}$ converges. We will formulate a one-sided multiplicative ergodic theorem which is based on Raghunathan’s version [Reference Raghunathan12].

Theorem 7. (Non-singular Oseledeč theorem)

Let ( $X,S,m$ ) be a probability space. Suppose that T is a non-singular transformation and $u:Z\times X\to M(r,{\mathbb {R}})$ is a measurable cocycle over T such that $\log ^{+}\|\Phi (1, \dot )\| \in L^{1}(X,S,m)$ . We set

$$ \begin{align*}B=\bigg\{(x,v)\in X \times {\mathbb{R}}^{r}: \frac{\log \|\Phi(n,x)v\|}{\sum_{i=0}^{n-1}\omega_{i}(x)}\,\, \mathrm{tends\ to\ a \ finite \ limit \ or\ }-\infty\bigg\}\end{align*} $$

and

$$ \begin{align*}X^{\prime}=\{x\in X|(x,v)\in B\,\, \mathrm{for\ all}\,\, v \in {\mathbb{R}}^{r}\}.\end{align*} $$

Then there is a subset Y of $X^{\prime }$ which has full measure and a sequence of functions $ \Lambda ^{1}(x) \leq \cdots \leq \Lambda ^{r}(x)$ (taking values in ${\mathbb {R}}\cup -\infty $ ) such that the following statements hold.

1. (i) Let $I=\{1=i_{1}<i_{2}<\cdots <i_{p}<i_{p+1}=r+1\}$ be $n+1$ tuples of integers. Define $Y(I)=\{x\in X^{\prime }|\Lambda ^{i}(x)=\Lambda ^{j}(x)$ for $ i_{q}\leq i,j<i_{q+1}$ and $\Lambda ^{i_{q}}(x)<\Lambda ^{i_{q+1}}(x)$ for all q with $1<q<p\}$ . Then for all $x\in Y(I),1<q<p$ ,

   $$ \begin{align*} \Sigma(I,q)(x)=\bigg\{v\in {\mathbb{R}}^{r}| \lim_{n\to \infty}\frac{\log \|\Phi(n,x)v\|}{\sum_{i=0}^{n-1}\omega_{i}(x)}\leq \Lambda^{i_q}(x)\bigg\} \end{align*} $$
   is a vector subspace of ${\mathbb {R}}^{r}$ with dimension $i^{q+1}-1$ .

2. (ii) If $\Sigma (I,0)=\{0\}$ then

   $$ \begin{align*} \lim_{n\to \infty}\frac{\log \|\Phi(n,x)v\|}{\sum_{i=0}^{n-1}\omega_{i}(x)}= \Lambda^{i_q}(x) \end{align*} $$
   for any vector $v \in \Sigma (I,q)(x) -\Sigma (I,q-1)(x)$ .

3. (iii) For $x\in Y$ , the sequence

   $$ \begin{align*} A(n,x)=(\Phi(n,x)^{*}\Phi(n,x))^{{1}/{2n}} \end{align*} $$
   converges to a matrix $A(x) \in M(r,{\mathbb {R}})$ . The eigenspace of $A(x)$ is the orthogonal complement of $\Sigma (I,q)(x)$ in $\Sigma (I,q+1)(x)$ corresponding to the eigenvalue $\exp \Lambda ^{i_{q+1}}$ .

Now, given $\varepsilon>0$ , there exists $N(\varepsilon ,x)$ such that for all $n>N$ ,

$$ \begin{align*} \| \Phi(1,T^{n}(x))\|< \exp \sum_{i=0}^{n-1}\omega_{i}(x)\varepsilon. \end{align*} $$

$\| \Phi (n,x)\|$ satisfies the cocycle identity: $\Phi (n+1,x)= \Phi (1,T^{n}(x)) \Phi (n,x)$ .

For a unit vector $v \in \Sigma (I,q,n)(x)$ ,

$$ \begin{align*}\|\Phi(n+1,x)v\|\leq \|\Phi(1,T^{n}(x))\|\| \Phi(n,x)v\|,\end{align*} $$

$$ \begin{align*}\|\Phi(n+1,x)v\|\leq \exp\bigg(\sum_{i=0}^{n-1}\omega_{i}(x)(\Lambda ^{i_{q}}+2\varepsilon)\bigg).\end{align*} $$

Choose a unit vector $v \in \Sigma (I,q,n)(x)$ , and let $v^{\prime } \in \Sigma (I,q,n+1)(x)$ be the orthogonal projection of v on $\Sigma (I,q,n+1)(x)$ . The orthogonal complement of $v^{\prime }$ in $\Sigma (I,q,n+1)(x)$ has the form $C_{n+1}^{-1}\sum _{i\geq _{q+1}-1}b_{i}e_{i}$ .

Lemma 3. Given $\varepsilon>0$ , there exists $N(\varepsilon ,x), x\in Y$ , with the following property. There is a unit vector $v \in \Sigma (I,q,n)(x)$ , for some number $b_{i}\in {\mathbb {R}}$ and $v^{\prime } \in \Sigma (I,q,n+1)(x)$ , such that

$$ \begin{align*} v=v^{\prime}+C_{n+1}^{-1}\sum_{i\geq i_{q+1}}b_{i}e_{i}. \end{align*} $$

Then $|b_{i}|<$ $\exp \{-\sum _{i=0}^{n-1}\omega _{i}(x)(\Lambda ^{i}-\Lambda ^{i_{q}}-\varepsilon )\}$ for $n\geq N$ .

Proof. We have $\|\Phi (n+1,x)v\|\leq \exp \sum _{i=0}^{n-1}\omega _{i}(x)(\Lambda ^{i_{q}}+2\varepsilon )$ by Lemma 2. Notice that

$$ \begin{align*} \|\Phi(n+1,x)v\|\geq \bigg\|\sum_{i\geq i_{q+1}}|b_{i}|\Phi(n+1,x)C_{n+1}^{-1}e_{i}\bigg\|. \end{align*} $$

Now $\|b_{i}\Phi (n+1,x)C_{n+1}^{-1}e_{i}\|\leq \|\sum _{i\geq i_{q+1}}b_{i}\Phi (n+1,x)C_{n+1}^{-1}e_{i}\|$ as $\Lambda ^{i}$ is non-decreasing. Hence

$$ \begin{align*} \|\Phi(n+1,x)v\|\geq \|b_{i}\Phi(n+1,x)C_{n+1}^{-1}e_{i}\|. \end{align*} $$

Now $\|\Phi (n+1,x)C_{n+1}^{-1}e_{i}\|$ is the ith eigenvalue, since

$$ \begin{align*} \|b_{i}\Phi(n+1,x)C_{n+1}^{-1}e_{i}\|=|b_{i}|\|L_{n+1}(D_{n+1})^{1/2}C_{n+1}C_{n+1}^{-1}e_{i}\|=|b_{i}|\delta_{i} (\Phi_{n+1}(x)). \end{align*} $$

Let $\Lambda ^{i}$ be the limit of ${\log \delta _{i} (\Phi _{n+1}(x))}/{\sum _{i=0}^{n}\omega _{i}(x)}$ . It follows from the above that

$$ \begin{align*} \|\Phi(n+1,x)v\|\geq|b_{i}| \exp\bigg(\bigg(\sum_{i=0}^{n}\omega_{i}(x)\bigg)(\Lambda^{i}-\varepsilon)\bigg). \end{align*} $$

Thus $\Lambda ^{i}$ is in a bounded interval which is greater than $\Lambda ^{i_{q}}$ , $0<\varepsilon <1$ . For large n, we may assume

$$ \begin{align*} \exp\bigg(\bigg(\sum_{i=0}^{n}\omega_{i}(x)\bigg)(\Lambda^{i}-\varepsilon)\bigg)\geq \exp\bigg(\bigg(\sum_{i=0}^{n-1}\omega_{i}(x)\bigg)(\Lambda^{i}-2\varepsilon)\bigg). \end{align*} $$

Thus

$$ \begin{align*} |b_{i}|\leq \exp -\bigg(\sum_{i=0}^{n-1}\omega_{i}(x)\bigg)(\Lambda^{i}-4\varepsilon-\Lambda ^{i_{q}}) \end{align*} $$

which completes the proof of the lemma.

Lemma 3 shows that a vector in $\Sigma (I,q,n)(x)$ can be combined with the projection on $\Sigma (I,q,n+1)(x)$ and the orthogonal complement of $\Sigma (I,q,n+1)(x)$ , that is,

$$ \begin{align*} C_{n}^{-1}\sum_{i=1}^{i_{q+1}-1}Ke_{i}=C_{n+1}^{-1}\sum_{j=1}^{r}b_{j}e_{j} \quad\mbox{and}\quad v_{n}=v_{m+1}^{\prime}+w_{m+1}. \end{align*} $$

Now $v_{m+1}^{\prime }$ is the orthogonal projection of $v_{n}$ onto $\Sigma (I,q,n+1)(x)$ and the norm of $v_{m+1}^{\prime }$ is given by

$$ \begin{align*} \|v_{m+1}^{\prime}\|=\bigg\|v_{n}-C_{n+1}^{-1}\sum_{i\geq i_{q+1}}b_{i}e_{i}\bigg\|. \end{align*} $$

Since $v_{n}$ is a unit vector, we have the following upper and lower bounds for $\|v_{m+1}^{\prime }\|$ :

$$ \begin{align*} 1-\bigg\|C_{n+1}^{-1}\sum_{i\geq i_{q+1}}b_{i}e_{i}\bigg\|\leq \|v_{m+1}^{\prime}\|\leq 1+\bigg\|C_{n+1}^{-1}\sum_{i\geq i_{q+1}}b_{i}e_{i}\bigg\|. \end{align*} $$

We want to show that $\|v_{n+1}-v_{n}\|\leq 2r \exp \{-(\sum _{i=0}^{n-1})\omega _{i}(x)(\Lambda ^{i_{q+1}}-\Lambda ^{i_{q}}-\varepsilon )\}$ .

Lemma 4. If $v_{n+1} \in \Sigma (I,q,n+1)(x)$ is a unit vector and $v_{n} \in \Sigma (I,q,n)(x)$ satisfies the conditions of Lemma 3, then

$$ \begin{align*} \|v_{n+1}-v_{n}\|\leq 2r \exp \bigg\{-\bigg(\sum_{i=0}^{n-1}\omega_{i}(x)\bigg)(\Lambda^{i_{q+1}}-\Lambda^{i_{q}}-\varepsilon)\bigg\}. \end{align*} $$

Proof.

$$ \begin{align*}\|v_{n+1}-v_{n}\|=\|v_{n}-v_{n+1}\|= \|v_{n}-v_{n+1}^{\prime}+v_{n+1}^{\prime}-v_{n+1}\|\end{align*} $$

and

$$ \begin{align*}\|v_{n}-v_{n+1}^{\prime}+v_{n+1}^{\prime}-v_{n+1}\|\leq\|v_{n}-v_{n+1}^{\prime}\|+\|v_{n+1}^{\prime}-v_{n+1}\|.\end{align*} $$

It follows that

$$ \begin{align*}\|v_{n}-v_{n+1}^{\prime}\|=\bigg\|C_{n+1}^{-1}\sum_{i\geq i_{q+1}}b_{i}e_{i}\bigg\|.\end{align*} $$

On the other hand, let $v_{n+1}^{\prime }=av_{n+1}$ , and we have $1-\|C_{n+1}^{-1}\sum _{i\geq i_{q+1}}b_{i}e_{i}\|\leq \|v_{m+1}^{\prime }\|\leq 1+\|C_{n+1}^{-1}\sum _{i\geq i_{q+1}}b_{i}e_{i}\|$ . Thus

$$ \begin{align*} 1-\bigg\|C_{n+1}^{-1}\sum_{i\geq i_{q+1}}b_{i}e_{i}\bigg\|\leq a\leq 1+\bigg\|C_{n+1}^{-1}\sum_{i\geq i_{q+1}}b_{i}e_{i}\bigg\|. \end{align*} $$

Now $\|v_{n+1}^{\prime }-v_{n+1}\| =\|(a-1)v_{n+1}\| $ , which is smaller than $\|C_{n+1}^{-1}\sum _{i\geq i_{q+1}}b_{i}e_{i}\|$ . We thus have

$$ \begin{align*} \|v_{n+1}-v_{n}\|\leq 2 \bigg\|C_{n+1}^{-1}\sum_{i\geq i_{q+1}}b_{i}e_{i}\bigg\|. \end{align*} $$

It follows that

$$ \begin{align*}\bigg\|C_{n+1}^{-1}\sum_{i\geq i_{q+1}}b_{i}e_{i}\bigg\| &\leq \sum_{i\geq i_{q+1}}\|b_{i}e_{i}\| \leq r|b_{i_{q+1}}| \\[3pt] &\leq r \exp \bigg\{-\bigg(\sum_{i=0}^{n-1}\omega_{i}(x)\bigg)(\Lambda^{i_{q+1}}-\Lambda^{i_{q}}-\varepsilon)\bigg\}.\end{align*} $$

This completes the proof.

We will show that $\|v_{n+k}-v_{n+l}\| $ is a Cauchy sequence. In fact, the sequence $\sum _{i=l}^{\infty } r \exp \{-(n+i)(\Lambda ^{i_{q+i}}-\Lambda ^{i_{q}}-\varepsilon )\}$ is the sum of a geometric series. Thus

$$ \begin{align*} &\|v_{n+k}-v_{n+l}\|\\[3pt] &\quad\leq \sum_{i=l}^{\infty}\|v_{n+k}-v_{n+l}\| \\[3pt] &\quad=2r \frac{1}{1-\exp \{-(\Lambda^{i_{q+1}}-\Lambda^{i_{q}}-\varepsilon)}\exp \bigg\{-\bigg(\sum_{i=0}^{n-1}\omega_{i}(x)+l\bigg)(\Lambda^{i_{q+1}}-\Lambda^{i_{q}}-\varepsilon)\bigg\}\\[3pt] &\quad=C \exp\bigg\{-\bigg(\sum_{i=0}^{n-1}\omega_{i}(x)+l\bigg)(\Lambda^{i_{q+1}}-\Lambda^{i_{q}}-\varepsilon)\bigg\}. \end{align*} $$

This shows that $v_{n}$ is a Cauchy sequence, as claimed.

Lemma 5. Let $\{v^{1}_{n},v^{2}_{n},\ldots ,v^{i}_{n}\}$ be a collection of vectors which is a basis for $\Sigma (I,q,n)(x)$ , where $ 0\leq i<i_{q+1}$ . The sequence $\{\Sigma (I,q,n)(x)\}$ has limit $\Sigma (I,q)(x)$ . Furthermore,

$$ \begin{align*}\|v_{n+k}-v_{n+l}\| \leq \sum_{i=l}^{k}\|v_{n+k}-v_{n+l}\| \leq \sum_{i=l}^{\infty}\|v_{n+k}-v_{n+l}\|.\end{align*} $$

Proof. By Lemma 4, we see easily that

$$ \begin{align*} \|v_{n+k}^{i}-v_{n+l}^{i}\|\leq C \exp\bigg\{-\bigg(\sum_{i=0}^{n-1}\omega_{i}(x)+\max(k,l)\bigg)(\Lambda^{i_{q+1}}-\Lambda^{i_{q}}-\varepsilon)\bigg\} \end{align*} $$

for $i_q \leq i < i_{q+1}$ . Moreover,

$$ \begin{align*} \|v_{n}^{i}-v^{i}\|\leq C \exp\bigg\{-\bigg(\sum_{i=0}^{n-1}\omega_{i}(x)\bigg)(\Lambda^{i_{q+1}}-\Lambda^{i_{q}}-\varepsilon)\bigg\} \end{align*} $$

where $v^{i}$ is the limit of $v_{n}^{i}$ . The sequence $\{v_{n}^{i}\}$ converges to a linearly independent set of vectors as $n\to \infty $ . $\Sigma (I,q,n)(x)$ is the space spanned by $\{v^{1}_{n},v^{2}_{n},\ldots ,v^{i}_{n}\}$ . Thus $ \Sigma (I,q,n)(x) \to \Sigma (I,q)(x)$ as $n\to \infty $ .

Writing the matrix $B=C_{n}C^{-1}_{n+k}$ , a vector in the space $\Sigma (I,q,n)(x)$ has the form $C_{n}^{-1}e_{i}$ which we can split into an orthogonal projection in $\Sigma (I,q,n+1)(x)$ and its orthogonal complement in $\Sigma (I,q,n+1)(x)^{\perp }$ . By Lemma 3, $C_{n}^{-1}e_{i}=C_{n+1}^{-1}\sum _{j=1}^{r}b_{j}e_{j}$ . Hence $|C_{n+1}C_{n}^{-1}|=|b_{j}|$ as $e_{i}$ is the standard basis. The inverse of $C_{n+1}C_{n}^{-1}$ is $C_{n}C_{n+1}^{-1}$ . Thus we have a similar decomposition of a vector in $\Sigma (I,q,n+1)(x)$ , namely,

$$ \begin{align*} v=v^{\prime}+C_{n}^{-1}\sum_{i\geq_{q+1}}b_{i*}e_{i}, \end{align*} $$

where $v^{\prime } \in \Sigma (I,q,n)(x)$ and $C_{n}^{-1}\sum _{i\geq _{q+1}}b_{i*}e_{i} \in \Sigma (I,q,n)(x)^{\perp } $ . We set $a_{i}= \exp (\Lambda ^{i}-i\varepsilon )$ , and note that $|C_{n}C_{n+1}^{-1}|=|b_{i*}|$ . It then follows that

$$ \begin{align*} |b_{i*}|\leq C \exp-\bigg(\sum_{i=0}^{n-1}\omega_{i}(x)\bigg)(\Lambda^{i}-\Lambda ^{i_{q}}-r\varepsilon), \end{align*} $$

since we have a cycle of length r.

Proof. Firstly, we can see that

$$ \begin{align*} \limsup ({\log \|\Phi (n,x)(v_{n}^{i})\|}/{\Omega (x)})=\lim _{n\to \infty }({\log \delta _{i}} {(\Phi (n,x))}/{\Omega (x)})=\Lambda ^{i}. \end{align*} $$

If $i<i_{q+1}$ , we have $\limsup ({\log \|\Phi (n,x)(v_{n}^{i})\|}/{\Omega (x)})\leq \Lambda ^{i_{q}}$ .

On the other hand, $ v^{i}-v_{n}^{i}=w+C_{n}^{-1}\sum _{i\geq _{q+1}}b_{i*}e_{i}$ . For $w\in \Sigma (I,q,n)(x)$ , $\limsup ({\log }$ ${\|\Phi (n,x)(w)\|}/{\Omega (x)})\leq \Lambda ^{i_{q}}$ . Hence

$$ \begin{align*} &\limsup\frac{\log \|\Phi(n,x)C_{n}^{-1}\sum_{i\geq i_{q+1}}b_{i*}e_{i}\|}{\sum_{i=0}^{n-1}\omega_{i}(x)}\\[3pt] &\quad\leq \limsup\frac{\log\{ C \exp-(\Omega(x))(\Lambda^{i}-\Lambda ^{i_{q}}-r\epsilon)\times \delta_{i} (\Phi(n,x))\}}{\Omega(x)}\\[3pt] &\quad=-\Lambda^{i}+\Lambda ^{i_{q}}+r\epsilon +\Lambda^{i}=\Lambda ^{i_{q}}+r\epsilon \end{align*} $$

for $i\geq i_{q+1}$ .

The triangle inequality implies that

$$ \begin{align*}\|\Phi(n,x)(v^{i})\|\leq \|\Phi(n,x)(v_{n}^{i})\|+\|\Phi(n,x)(v^{i}-v_{n}^{i})\| \leq 2 \exp \bigg(\sum_{i=0}^{n-1}\omega_{i}(x)\bigg)(\Lambda ^{i_{q}}+\varepsilon), \end{align*} $$

and thus $\limsup ({\log \|\Phi (n,x)(v^{i})\|}/{\sum _{i=0}^{n-1}\omega _{i}(x)})\leq \Lambda ^{i_{q}}$ .

Lemma 7. If the vector $v^{i}$ is not in $\Sigma (I,q-1)(x)$ , for large n, the projection $v^{i\prime }$ belongs to $\Sigma (I,q-1,n)(x)$ with $\|v^{i\prime }\|>c>0$ . Then

$$ \begin{align*} \liminf\frac{\log\|\Phi(n,x)(v^{i})\|}{\sum_{i=0}^{n-1}\omega_{i}(x)} \geq \Lambda^{i_q}(x). \end{align*} $$

Proof. This proof is quite straightforward. We take a unit vector $v^{i}$ which is not in $\Sigma (I,q-1)(x)$ . There is a $\delta \in V$ such that $v^{i}+\delta \in \Sigma (I,q-1)(x) $ . When n is large enough, the vector $v^{i}_{n}$ has projection $v^{i\prime }$ in $\Sigma (I,q-1,n)(x)$ and orthogonal complement ${v^i}'' \in \Sigma (I,q-1,n)(x)^{\perp } $ . We take the difference $\|v^{i}-v^{i\prime }\|\geq {\delta }/{2}$ , obtaining

$$ \begin{align*} \|\Phi(n,x)(v^{i})\|& \approx \|\Phi(n,x)v^{i}_{n}\|\\[3pt] &\geq \|\Phi(n,x)(v^{i}-v^{i\prime})\|\\[3pt] &\geq \frac{\delta}{2}\exp \bigg(\sum_{i=0}^{n-1}\omega_{i}(x)\bigg)(\Lambda^{i_q}(x)-\varepsilon).\\[-3.8pc] \end{align*} $$

Combining Lemmas 6 and 7, we can conclude that

$$ \begin{align*} \lim ({\log \|\Phi (n,x)(v^{i})\|}/ \sum _{i=0}^{n-1} \omega _{i}(x)) =\Lambda ^{i_q}(x),\end{align*} $$

for $v_{i} \in \Sigma (I,q)(x)\setminus \Sigma (I,q-1)(x)$ . We now see that the eigenspace is $C_{n}^{-1}e_{i} \to C^{-1}e_{i}$ and the eigenvalue is $\Lambda ^{k}(x)=\lim _{n\to \infty }({ \log \delta _{k} (\Phi (n,x))}/ {\sum _{i=0}^{n-1}\omega _{i}(x)})$ , so that the limit matrix $A(x) =\lim _{n\to \infty } (\Phi (n,x)^{*}\Phi (n,x))^{{1}/{2n}}$ exists.